IN  MEMORIAM 
FLOR1AN  CAJORI 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

Microsoft  Corporation 


http://www.archive.org/details/elementarycourseOOhackrich 


I   t 


ELEMENTARY  COURSE 


GEOMETRY 


FOB    THE    USE    OF 


SCHOOLS  AND  COLLEGES. 


.IX 


CHARLES  \Y.  KACKLEY,  S.T.I), 
if 

SOR    OF    MATHEMATICS    AND 'ASTRONOMY  IN   COLUMBIA   CC 
AND  AUTHOR   OF   A   "TREATISE   ON  ALGEBRA,"   ETC. 


NEW    YORK: 

HARPER  &   BROTHERS,   PUBLISHERS, 
82   CLIFF   STREET. 

1  84  7. 


H27 


Entered,  according  to  Act  of  Congress,  in  the  year  one  thousand 
eight  hundred  and  forty-seven,  by 

Harper  &.  Brothers, 

in  the  Clerk's  Office  of  the  District  Court  of  the  Southern  District 
of  New  York. 


CAJbm 


PREFACE. 


The  materials  for  the  following  work,  like  those 
of  the  author's  treatise  on  Algebra,  have  been  drawn 
from  the  latest  and  best  foreign  sources,  and  from  the 
results  of  a  varied  experience  of  near  twenty  years 
as  an  instructor,  commencing  at  the  United  States 
Military  Academy. 

The  definitions  of  angles  and  parallel  lines,  upon 
which  so  much  depends,  will  be  found  quite  different 
from  those  in  ordinary  use  ;  yet  it  is  believed  that  no 
others  hitherto  suggested  are  so  direct  and  distinct, 
so  free  from  metaphysical  objections,  or  so  easily  ap- 
prehended by  the  learner ;  and  none,  certainly,  are 
productive  of  so  much  simplicity,  generality,  and 
brevity  in  the  depending  demonstrations. 

The  treatment  of  proportions  as  equalities  of  ra- 
tios, it  is  thought,  will  give  greater  clearness  to  the 
proof  of  those  propositions  in  which  they  are  used. 

Great  care  has  been  taken  to  remedy  all  the  little 
imperfections  of  demonstration  in  older  treatises,  and 
to  supply  some  propositions  which  have  been  here- 
tofore unaccountably  omitted. 

The  infinitesimal  system  has  been  adopted  without 

91882/' 


IV  PREFACE. 

hesitation,  and  to  an  extent  somewhat  unprecedented. 
The  usual  expedients  for  avoiding  this,  result  in  te- 
dious methods,  involving  the  same  principle,  only  un- 
der a  more  covert  form.  The  idea  of  the  infinite  is 
certainly  a  simple  idea,  as  natural  to  the  mind  as  any 
other,  and  even  an  antecedent  condition  of  the  idea 
of  the  finite. 

A  peculiar  feature  of  the  work  will  be  observed  in 
the  "  Exercises,"  which  occur  at  intervals,  commenc- 
ing immediately  after  the  Axioms.  These  are  intend- 
ed to  develop  the  original  powers  of  the  learner,  and 
to  bring  into  play  his  inventive  faculties,  the  ordinary 
text  tasking  the  powers  of  perception  alone.  The 
Exercises  are  so  arranged  as  to  make  the  progress 
from  the  easy  to  the  more  difficult  so  gradual  that 
they  will  be  found  to  excite  a  lively  interest  even  in 
students  of  moderate  capacity. 

They  will  be  especially  convenient  in  the  instruc- 
tion of  large  classes,  the  members  of  which  may  all 
pursue  the  text,  while  the  exercises  upon  it  will  afford 
scope  for  the  students  of  greatest  ability. 

The  appendices  will  be  found  to  contain  some  re- 
cent and  elegant  improvements.  They  leave  much 
to  be  done  by  the  learner,  it  being  supposed  that  none 
will  be  likely  to  attempt  them  except  such  as  have 
some  taste  and  talent  for  geometry.  The  previous 
exercises  will  have  furnished  the  skill  requisite  to 
master  this  part  of  the  work  with  facility. 

The  work  might  have  been  arranged  in  more  ele- 


PRE1 


gant  form  by  a  rigid  classification  of  subjects,  all  the 
theorems  relating  to  a  particular  class  of  magnitudes 
being  given  together,  after  the  manner  of  some  of  the 
latest  and  best  French  and  German  treatises.  This 
arrangement,  though  in  the  main  preserved,  has  been 
occasionally  departed  from,  for  the  sake  of  rendering 
a  knowledge  of  the  whole  subject  most  easy  of  ac- 
quisition by  the  student. 

By  omitting  the  fine  print  in  the  volume,  the  stu- 
dent will  obtain  a  very  short  course  of  geometry, 
but  one  fully  adequate  as  a  preparation  for  the  study 
of  all  the  higher  branches  of  mathematics,  while  the 
whole  work  contains,  probably,  the  most  complete 
system  of  purely  elementary  geometry  to  be  found  rn 
any  single  treatise  in  any  language. 
A2 


CONTENTS, 


PLANE  GEOMETRY 

Definitions 

Axioms 

>es  upon  the  right  line  and  angle 
Theorems  relating  to  identical  triangles 

"  "  isosceles  triangles 

Exercises  upon  the  foregoing  theorems 
Theorems  upon  intersecting  lines 
Relative  magnitudes  of  angles  of  triangles 
Theorems  upon  secant  lines  and  parallels 
Theorems  relating  to  parallels     . 
Relations  of  outward  and  inward  angles  of  triangles 
Sum  of  the  angles  of  triangles  and  other  polygons 

Theory  of  perpendiculars 

Properties  of  parallelograms        .... 
Relations  of  parallelograms  to  triangles 

"  i*  trapezoids     . 

"        of  the  squares  of  the  sides  of  triangles 
Exercises  on  the  foregoing  .... 

Exposition  of  the  nature  of  analysis  and  synthesis 
Specimen  of  the  analysis  of  a  problem 

THE  CIRCLE  AND  ITS  COMBINATIONS  WITH  RIGHT  LINES. 

Theorems  relating  to  chords  in  a  circle 

"              "            tangents  to  a  circle     . 
Measures  of  angles  in  various  positions  in  a  circle 
Theorems  relating  to  secants  of  a  circle 
Equiangular  triangles           ..... 
Exercises  upon  the  preceding  theorems 
Numerical  problems 

RATIOS  AND  PROPORTIONS. 

Definitions  ........ 

Theorems  relating  to  ratio  and  proportion  .         .         .         .         .52 

■-  of  triangles  and  parallelograms 56 

Theory  of  the  measure  of  parallelograms  and  triangles        .        .    59 


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50 


Vlll  CONTENTS. 

Pago 

Division  of  the  sides  of  triangles  in  various  proportions       .         .     60 

Theory  of  similar  triangles 62 

"  "      polygons 65 

"  **  "        inscribed  in  circles    .         .         .         .67 

Ratios  of  the  elements  of  a  circle        ......     69 

Area  of  the  circle 70 

Exercises  upon  the  circle    .         .         .         .         .         •         •         .71 

PROBLEMS  IN  PLANE  GEOMETRY. 

Problems  relating  to  perpendiculars 74 

"  "        to  the  division  and  construction  of  angles        .     76 

"  "         to  parallels 77 

"  "         to  the  construction  of  triangles        .         .         .78 

"  "  "  "  squares  and  rectangles      79 

"  "         to  equal  figures 81 

"  "         to  the  describing  of  circles      .         .         .         .82 

u  "         to  the  drawing  of  tangents      .         .         .         .83 

Miscellaneous      ..........     84 

Problems  upon  regular  figures 88 

"  "      similar  figures 92 

General  note  upon  the  method  of  solution  of  problems    .         .         .93 
Miscellaneous  exercises  in  plane  geometiy  .         .         .         .100 

APPENDIX  I. 

ISOPERIMETRY. 

Of  triangles 1 

Of  polygons 2 

Of  plane  figures 4 

APPENDIX  II. 

Centers  of  symmetry 1 

Of  axes  of  symmetry  .........  1 

Of  diameters       .         .........  1 

Center  of  mean  distances 1 

Centers  of  similitude 2 

"                "           in  circles     .......  4 

Radical  axis  and  radical  center 5 

Conjugate  points,  poles,  and  polar  lines 6 

GEOMETRY  OF  PLANES. 

Definitions 1 

Sections  of  planes  with  one  another    ......       2 

Conditions  which  fix  the  position  of  a  plane        ....       3 


CONTENTS.  IX 

Page 

Theory  of  perpendiculars  to  planes 4 

"         parallel  planes  and  lines 8 

"         perpendicular  planes 12 

Exercises  upon  planes 14 

POLYHEDRAL  ANGLES. 

Definitions  ...........  1 

Relat  ions  of  the  plane  angles  of  polyhedral  angles  1 

"             "     diedral  angles 3 

Symmetry  in  polyhedral  angles 5 

Exercises  upon  polyhedral  angles         .         .     '  .         .         .        .7 

SOLID  GEOMETRY. 

Definitions  ...........  1 

Til  i  positions  relating  to  sections  of  prisms  and  cylinders     .         .  4 

Ratios  of  parallelopipedons 6 

Theory  of  the  measure  of  parallelopipedons         ....  8 

Tangent  planes  to  cylinders,  and  development  of  their  surface   .  8 

Ratios  of  similar  prisms  and  cylinders 9 

Propositions  relating  to  sections  of  pyramids  and  cones       .         .  9 

Relations  of  pyramids  and  prisms 12 

Tangent  planes  to  cones,  and  development  of  their  surface          .  13 

-os  in  solid  geometry ]3 

r'ru>tum  of  the  pyramid  and  cone 1G 

Ratio  of  sphere  and  circumscribing  cylinder        .         .         ,         .18 

Mea>ure  of  a  spherical  segment 19 

SPHERICAL  GEOMETRY. 

Definitions 1 

Sections  of  the  sphere 2 

Propositions  relating  to  the  poles  and  secondaries  of  circles         .  4 

Tangent  plane  to  the  sphere 7 

Various  measures  of  spherical  angles 8 

Polar  or  supplemental  triangles 9 

Relations  of  the  sides  and  angles  of  spherical  triangles        .         .  11 

Theorems  relating  to  the  identity  and  symmetry  of  sph.  triangles  12 

Dim  its  to  the  sum  of  the  angles  of  a  spherical  triangle        .         .  14 
Ratio  of  a  lune  to  the  surface  of  the  sphere          .         .         .         .15 

"        spherical  triangle  to  the  surface  of  the  sphere      .         .  16 

Exercises  in  spherical  geometry  .......  17 

APPENDIX  III. 
Relation  of  ratio  of  an  arc  to  the  quadrant  with  the  ratio  of  an 

arc  to  its  radius 1 


CONTENTS. 


Expression  for  an  angle  in  terms  of  unit  of  arc  and  unit  of  length  1 

Relation  of  arcs  having  a  common  chord  and  different  radii        .  1 

The  shortest  path  on  the  surface  of  the  sphere     ....  2 

APPENDIX  IV. 

ISOPERIMETRY  ON   THE   SPHERE. 

Of  spherical  triangles 1 


Of  spherical  polygons 


APPENDIX  V. 


SYMMETRY   IN   SPACE. 

Symmetry  of  position 1 

"          relative  to  an  axis 1 

"          with  reference  to  a  plane 2 

Diametral  planes 4 

Center  of  mean  distances    ........  5 

Of  centers  of  similitude        ........  6 

Centers  of  similitude  of  spheres 6 

REGULAR  POLYHEDRONS. 

Proof  that  there  can  be  but  live 7 

Construction  of  the  regular  tetrahedron 8 

"             "             "       hexahedron 8 

"             "             "        octahedron 8 

"             "             "       icosahedron      .....  8 

"             "             "       dodecahedron           ....  9 

Relation  of  the  number  of  vertices,  edges,  and  faces  of  a  polyhedron  1 0 

MENSURATION. 


MENSURATION    OF    PLANES. 

Area  of  a  parallelogram 
Examples    . 
Area  of  a  triangle 
Examples    . 
Area  of  a  trapezoid 
Examples    . 

Area  of  a  trapezium,  and  examples 
Of  an  irregular  polygon,  and  example 
Of  a  regular  polygon    . 
Examples  and  table     . 
Demonstration  of  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter 


CONTENTS. 


XI 


Examples 9 

To  line  1  iln*  length  of  an  arc,  and  examples  .         .         .         .10 

of  a  circle 10 

Kxamples "11 

Area  of  a  circular  ring,  and  examples 11 

Area  of  the  sector  of  a  circle,  and  examples         .         .         .         .12 

"  segment  of  a  circle,  and  examples     .         .         .         .12 

Table  of  areas  of  segments 14 

Area  of  long  irregular  figures      .......  14 

Kxamples 15 

MENSURATION  OF  SOLIDS. 

Superficies  of  a  prism 1 

Examples 2 

Superficies  of  an  irregular  polyhedron 2 

"          of  a  regular  polyhedron 2 

"         of  a  pyramid  or  cone,  and  examples   ....  2 

"          of  a  frustum,  and  examples 3 

Volume  of  a  prism  or  cylinder 3 

Examples 4 

Volume  of  a  pyramid  or  cone,  and  examples       ....  4 

"       of  a  frustum,  and  examples 5 

Surface  of  a  sphere  or  segment 5 

Examples .  6 

Surface  of  a  lune,  volume  of  a  wedge,  surface  of  a  spherical  tri- 
angle, and  of  a  spherical  polygon,  and  examples      ...  7 
Volume  of  a  sphere,  and  examples      ......  8 

u       of  a  spherical  sector,  and  examples         ....  8 

"                    •*          segment,  and  examples     ....  9 

Exercises  in  mensuration 9 


GEOMETRY. 


DEFINITIONS. 

Geometry  is  the  science  of  position  and  extension. 

1.  A  Point  is  position  without  magnitude  or  di- 
mensions. It  has  neither  length,  breadth,  nor  thick- 
ness. 

2.  A  Line  has  one  dimension  only,  length. 

3.  A  Surface  or  Superficies  has  extension  in  two 
dimensions,  length  and  breadth ;  but  is  without  thick- 
ness.* 

4.  A  Body  or  Solid  has  three  dimensions,  length, 
breadth,  and  depth  or  thickness. 

5.  A  Right  Line,  or  Straight  Line,  is  A  B 
one  which  has  every  where  the  same  " 
direction. 

When  the  term  Line  is  used  in  this  work  without 
an  adjective,  a  Right  Line  is  understood. 

A  line  is  designated  by  two  letters  placed  upon  it. 
Thus  we  say  the  line  AB. 

6.  A  Broken  Line  is  one  which 
changes  its  direction  at  intervals. 

7.  A  Curve,  or  Curve  Line,  is  one 
which  is  continually  changing  its  di- 
rection. 

8.  Parallel  Lines  are  those  which  

have  the  same  direction. f  

*  A  surface  may  be  boundless,  and  a  line  interminable. 

1  Parallel  lines  are  sometimes  said  to  meet  at  an  infinite  distanre; 
in  other  words,  they  never  meet.  This  follows  evidently  from  the 
definition. 

The  case  where  one  line  is  the  prolongation  of  another,  or  others, 
which  seeim  t->  he  euthneed  in  this  definition,  is  to  he  excluded; 
for  those  lines,  in  lie-  u:uv-t  neted  WUe  of*  tin-  torn,  form  one  ami 
the  lame  Straight  line.  Or,  when  two  parallel  lines  coincide,  they 
become  one  and  the  sam  •  straight  line. 

A 


2  GEOMETRY. 

9.  Olio  line  is  Perpendicular  to  an- 
other when  the  first  inclines  not  more 
toward  the  second  on  the  one  side  than 
on  the  other.  _ 

10.  An  Angle  is  the  difference  of  direc- 
tion of  two  lines.* 

The  point  where  the  two  lines  meet  is 
called  the  vertex  of  the  angle. 


11.  Angles  are  Right  or  Oblique. 

12.  A  Right  Angle  is  that  which  is  made  by  l 
one  line  perpendicular  to  another. 

Or,  when  the  angles  on  either  side  of  one 
line  meeting  another  are  equal,  they  are 
right  angles. 

13.  Oblique  Angles  are  either  Acute  or  Obtuse. 

14.  An  Acute  Angle  is  less  than  a  right 
angle. 

15.  An   Obtuse  Angle  is  greater 
than  a  right  angle. 

*  When  one  line,  having  coincided  with  another,  begins  to  move 
round  the  point  at  one  extremity,  it  begins  to  have  a  different  direc- 
tion, and  the  amount  of  this  difference  depends  upon  the  amount  of 
the  movement,  which  is  evidently  measured  by  the  portion  of  the 
circumference  described  by  the  other  extremity.  The  length  of  this 
portion  is  usually  expressed  in  degrees,  each  degree  being  the  3^ th 
part  of  the  whole  circumference. 

It  is  immaterial  whether  the  revolving  line  be  longer  or  shorter, 
when  it  has  attained  the  same  position  with  respect  to  the  stationary 
line,  or  the  same  difference  of  direction  from  it,  the  portion  of  cir- 
cumference described  will  contain  the  same  number  of  degrees  in 
both  circumferences ;  each  degree  in  the  smaller  circumference  being 
smaller,  since  it  is  the  360th  part  of  its  own  circumference. 

Parallel  lines  having  no  difference  of  direction,  the  angle  which 
they  make  with  each  other  is  zero  or  0°. 

Fractions  of  a  degree  are  expressed  usually  in  minutes  and  seconds, 
a  minute  being  the  60th  part  of  a  degree,  and  a  second  the  60th  part 
of  a  minute.  This  is  called  the  sexagesimal  measurement  of  angles,  or 
division  of  the  circumference.  Another  mode  of  division  sometimes 
used,  is  of  the  whole  circumference  into  four  parts  called  quadrants, 
the  quadrant  into  100  parts  called  grades,  or  centesimal  degrees, 
each  grade  into  100  centesimal  minutes,  and  each  minute  into  100 
centesimal  seconds.  This  is  called  the  centesimal  division  of  the  cir- 
cumference. To  convert  one  kind  of  degree  into  the  other,  it  is  only 
necessary  to  observe  that  a  grade  is  0-9  of  a  degree. 


DEFINITIONS. 

All  angle  is  named  from  the  letter  at  its  vertex. 
Thus    we   say   the   angle    A.     When,    however, 
there  are  two  angles  whose  vertices  are  at  the 
same   point,   this  method  would   be  ambiguous. 
It  is  necessary,  then,  to  designate  the  angle  to  be  -A 
pointed  out  by  three  letters,  naming  the  one  at 
the  vertex  always  in  the  middle.     Thus,  the 
angle  formed  by  the  two  lines  CB  and  CE  is 
called  the  angle  BCE,  or  ECB  ;  and  the  angle 
formed   by  the  two  lines  CE  %nd  CD  is  called 
the  angle  ECD,  or  DCE. 

Angles  are  susceptible  of  addition,  subtraction,  and  multiplication. 
Thus  the  angle  BCD  =  BCE  +  ECD. 

16.  Superficies  are  either  Plane  or  Curved. 

17.  A  Plane  Superficies,  or  a  Plane,  is  that  which  is 
straight  in  every  direction,  or  with  which  a  right  line, 
joining  any  two  points  of  it,  will  coincide  throughout 
the  length  of  the  line.     But  if  not,  it  is  curved. 

18.  More  accurately,  a  Curve  Surface  is  one  of 
which  the  section  made  by  some  plane  cutting  it  is 
a  curve.* 

19.  A  Plane  Figure  is  a  portion  of  a  plane,  bound- 
ed either  by -right  lines  or  curves. 

20.  Plane  figures  that  are  bounded  by  right  lines 
are  called  Polygons,  and  have  names  according  to 
the  number  of  their  sides,  or  of  their  angles ;  the 
number  of  sides  and  angles  being  the  same.  The 
least  number  of  sides  requisite  to  form  a  polygon  is 
three. 

21.  A  Polygon  of  three  sides  and  three  angles  is 
called  a  Triangle.  And  it  receives  particular  de- 
nominations from  the  relations  of  its  sides  and  an- 
gles. 


22.  An  Equilateral  Triangle  is  one  the 
three  sides  of  which  are  all  equal. 

*  A  curve  surface  may  or  may  not  be  straight  in  certain  directions. 
See  the  cone  and  cylinder,  toward  the  end  of  the  volume. 


GEOMETRY. 


23.  An  Isosceles  Triangle  is  one  which  has 
two  sides  equal. 


24.  A  Scalene  Triangle  is  one  whose  three  sides  are 
all  unequal. 

25.  A  Right-angled  Triangle  is  a  tri- 
angle having  one  right  angler 

It  will  be  shown  hereafter  that  no  triangle  can  have  more  than  ono 
right  angle,  or  more  than  one  obtuse  angle. 

26.  Other  triangles  are  Oblique-angled,  and  are 
either  obtuse  or  acute. 

27.  An  Obtuse-angled  Triangle  has  one  obtuse  angle. 

28.  An  Acute-angled  Triangle  has  its  three  angles 
acute. 

29.  A  figure  of  Four  sides  and  angles  is  called  a 
Quadrangle,  or  a  Quadrilateral. 

30.  A  Parallelogram  is  a  quadrilateral  which  has 
both  its  pairs  of  opposite  sides  parallel.  And  it 
takes  the  following  particular  names,  viz.,  Rectangle, 
Square,  Rhombus,  Rhomboid. 

31.  A  Rectangle  is  a  right-angled  par 
allelogram. 


32.  A  Square  is  an  equilateral  rectangle. 


33.  A  Rhomboid  is  an  oblique-an- 
gled parallelogram. 

34.  A   Rhombus    is   an   equilateral 
rhomboid. 

35.  A  Trapezium  is  a  quadrilateral  which  has  not 
its  opposite  sides  parallel. 

36.  A  Trapezoid  is  a  quadrilateral 
which  has  only  one  pair  of  opposite 
sides  parallel. 


DEFINITIONS.  o 

37.  A  Pentagon  is  a  polygon  of  five  sides ;  a  Hex- 
agon is  one  of  six  sides  ;  a  Heptagon,  one  of  seven  ; 
an  Octagon,  one  of  eight ;  a  Nonagon,  one  of  nine  ; 
a  Decagon,  one  of  ten  ;  an  Undeeagon,  one  of  eleven  ; 
and  a  Dodecagon,  one  of  twelve  sides. 

The  Perimeter  of  a  polygon  is  the  sum  of  its  bound- 
ing lines. 

A  Convex  Polygon  is  one  the  perimeter  of  which 
can  be  intersected  by  a  straight  line  in  but  two 
points. 

38.  A  Polygon  is  Equilateral  when  all  its  sides  are 
equal ;  and  it  is  Equiangular  when  all  its  angles  are 
equal.  A  Regular  Polygon  is  one  which  is  both  equi- 
angular and  equilateral. 

39.  An  Equilateral  Triangle  is  a  regular  polygon 
of  three  sides,  and  the  square  is  one  of  four;  the  for- 
mer being  also  called  a  trigon,  and  the  latter  a  tetra- 
gon. 


40.  A  Diagonal  is  a  line  joining  any 
two  angles  of  a  polygon  not  adjacent. 


41.  A  Circle  is  a  plane  figure  bounded  by 
a  curve  line,  called  the  Circumference,  every 
point  of  which  is  equidistant  from  a  certain 
point  within,  called  the  Center. 

42.  The  Radius  of  a  circle  is  a  line  drawn 
from  the  center  to  the  circumference. 

43.  The  Diameter  of  a  circle  is  a  line 
drawn  through  the  center,  and  terminating 
both  ways  at  the  circumference. 

44.  An  Arc  of  a  circle  is  any  part  of  the 
circumference.*  • 

*  It  will  be  shown  hereafter  that  the  circumference  of  a  circle  may 
be  obtained  by  multiplying  the  radius  by  6:28:3:2:  hi  order  to  obtain 
the  absolute  length  of  any  arc  given  in  degrees  and  parts  of  a  degree, 
or  grades  and  parts,  it  is  necessary  to  ascertain  what  fraction  of  a 


D  GEOMETRY. 

45.  A  Chord  is  a  right  line  joining  the  ex- 
tremities of  an  arc. 

46.  A  Segment  is  any  part  of  a  circle 
bounded  by  an  arc  and  its  chord. 

47.  A  Semicircle  is  half  the  circle,  or  a 
segment  cut  off  by  a  diameter.  A  Semicir- 
cumference  is  half  the  circumference. 

48.  A  Sector  is  a  part  of  a  circle  which  is 
bounded  by  an  arc,  and  two  radii. 

Note. — A  sector  is  a  surface,  as  is  also  a 
segment. 

49.  A  Quadrant,  or  Quarter  of  a  circle,  is 
a  sector  having  a  quarter  of  the  circumfe- 
rence for  its  arc,  its  two  radii  being  perpen- 
dicular to  each  other.  A  quarter  of  the  cir- 
cumference is  also  called  a  Quadrant. 

Note. — A  semicircle  contains  180  degrees,  and  a 
quadrant  90  degrees. 

50.  Concentric  Circles  are  those  which  have  the 
same  center. 

51.  Circles  are  said  to  be  Eccentric  with  respect 
to  one  another  when  they  have  not  the  same  center. 
In  this  case,  the  one  circumference  may  be,  with  re- 
spect to  the  other,  Exterior,  Interior,  Tangent  Exter- 
nally, Tangent  Internally,  or,  finally,  the  two  circum- 
ferences may  intersect. 

52.  An  Angle  in  a  Segment  is 
that  which  is  contained  by  two 
lines,  drawn  from  any  point  in  the 
arc  of  the  segment,  to  the  two 
extremities  of  that  arc.  Thus  A 
and  D  are  both  angles  in  the  seg- 
ment BADC.  They  are  also  call- 
ed inscribed  angles,  and  are  said 
to  be  inscribed  in  the  segment. 

circumference  the  arc  is,  by  reducing  360°  to  the  lowest  denomina- 
tion in  the  given  arc  for  a  denominator,  and  the  degrees,  &c,  of  the 
arc  to  the  same  denomination,  for  a  numerator,  then  to  multiply  this 
fraction  by  the  product  of  the  radius  and  the  number  6-283:2. 


DEFINITIONS. 


are 


those 


53.  An  Angle  on  a  Segment,  or  an  Arc,  is  that  which 
is  contained  by  two  lines,  drawn  from  any  point  in 
the  opposite  part  of  the  circumference  to  the  extrem- 
ities of  the  arc,  and  containing  the  arc  between  them. 
Thus  A  and  D  (in  the  last  figure)  are  both  angles  upon 
the  arc  BEC. 

54.  An  Angle  at  the  Center  is  one  whose 
vertex  is  at  the  center  of  the  circle.  An 
Eccentric  Angle  is  one  whose  vertex  is  not  at 
the  center.  An  Angle  at  the  Circumference 
is  one  whose  vertex  is  in  the  circumfe- 
rence. This  last  is  also  called  an  Inscribed 
angle. 

55.  Similar  arcs,  in  different  circles, 
which  subtend  equal  angles  at  the  center. 

50.  A  right  line  is  a  Tangent  to  a  circle, 
or  touches  it,  when  it  has  but  one  point  in 
common  with  the  circle. 


57.  Two  circles  Touch  each  other  when 
they  have  but  one  point  common,  or  when 
they  have  a  common  tangent. 

58.  A  right-lined  figure  is  Inscribed  in  a 
circle,  or  the  circle  Circumscribes  the  fig- 
ure, when  all  the  angular  points  of  the  fig- 
ure are  in  the  circumference  of  the  circle. 

59.  A  right-lined  figure  Circumscribes  a 
circle,  or  the  circle  is  Inscribed  in  the  figure, 
when  all  the  sides  of  the  figure  touch  the 
circumference  of  the  circle. 

60.  One  right-lined  figure  is  inscribed  in 
another,  or  the  latter  circumscribes  the 
former,  when  all  the  angular  points  of  the 
former  are  placed  in  the  sides  of  the  lat- 
ter. 


61.  A  Secant  is  a  line  that  cuts  a  cir-     f 
cle,  lying  partly  within  and  partly  with-     ! 

out  it.  V- 


8  GEOMETRY. 

62.  The  Altitude  of  a  triangle  is  a 
perpendicular  let  fall  from  the  ver- 
tex of  either  angle  upon  the  opposite 
side,  called  the  base. 

(J.'*.  In  a  right-angled  triangle,  the  side  opposite 
the  right  angle  is  called  the  Hypothenuse  ;  and  the 
other  two  sides  are  called  the  Legs,  and  sometimes 
the  Base  and  Perpendicular. 

04.  The  altitude  of  a  parallelogram 
or  trapezoid  is  the  perpendicular  dis- 
tance between  the  parallel  sides. 

The  bases  of  a  trapezoid  are  the  par 
allel  sides. 


65.  Two  triangles,  or  other  right-lined  figures,  are 
said  to  be  mutually  equilateral  when  all  the  sides  of 
the  one  are  equal  to  the  corresponding  sides  of  the 
other,  each  to  each ;  and  they  are  said  to  be  mutu- 
ally equiangular  when  the  angles  of  the  one  are  re- 
spectively equal  to  those  of  the  other. 

66.  Identical  polygons  are  such  as  are  both  mutu- 
ally equilateral  and  equiangular,  or  that  have  all  the 
sides  and  all  the  angles  of  the  one  respectively  equal 
to  all  the  sides  and  all  the  angles  of  the  other,  each  to 
each  ;  so  that  if  the  one  figure  were  applied  to,  or  laid 
upon  the  other,  all  the  sides  of  the  one  would  exactly 
fall  upon  and  cover  all  the  sides  of  the  other  ;  the  two 
becoming,  as  it  were,  but  one  and  the  same  figure. 

67.  Similar  polygons  are  of  the  same  shape,  but  not 
the  same  size ;  they  have  all  the  angles  of  the  one 
equal  to  all  the  angles  of  the  other,  each  to  each,  and 
the  corresponding  or  homologous  sides,  as  they  are 
called,  proportional.*    The  homologous  sides  are  those 

*  Perhaps  it  will  be  a  little  plainer  to  sav  that  the  homologous  sides 
in  the  two  figures  have  the  same  ratio.  Thus,  if  the  first  side  in  the 
one  figure  (beginning  in  both  at  the  sides  adjacent  equal  ansles)  be 
three  times  a9  great  as  the  first  side  in  the  other,  the  second  side  in 
the  first  figure  will  be  three  times  as  great  as  the  second  side  in  the 
other  figure,  and  so  on. 

The  ratio  of  the  corresponding  sides  of  the  polygon  is  called  the 
ratio  of  similitude. 


AXIOMS.  9 

similarly  situated,  or  those  adjacent  equal  angles,  or, 
in  triangles,  those  opposite  equal  angles. 

68.  A  Proposition  is  something  which  is  either 
proposed  to  be  done,  or  to  be  demonstrated,  and  is 
either  a  problem  or  a  theorem. 

89.  A  Problem  is  something  proposed  to  be  done. 

70.  A  Theorem  is  a  truth  proposed  to  be  demon- 
strated. 

71.  A  Hypothesis  is  a  supposition  made  in  theenun- 
ciation  of  a.  proposition,  or  in  the  course  of  a  demon- 
stration. 

72.  A  Lemma  is  something  which  is  premised,  or 
demonstrated,  in  order  to  render  what  follows  more 
easy. 

73.  A  Corollary  is  a  consequent  truth,  gained  im- 
mediately from  some  preceding  truth  or  demonstra- 
tion. 

74.  A  Scholium  is  a  remark  or  observation  made 
upon  something  going  before  it,  and  may  require  a 
demonstration  or  may  not. 

Axioms. 

1.  Things  which  are  equal  to  the  same  thing  are 
equal  to  one  another. 

2.  When  equals  are  added  to  equals,  the  wholes 
are  equal. 

3.  When  equals  are  taken  from  equals,  the  remain- 
ders are  equal. 

4.  When  equals  are  added  to  unequals,  the  wholes 
are  unequal. 

'  5.  When  equals  are  taken  from  unequals,  the  re- 
mainders are  unequal. 

6.  Things  which  are  double  the  same  thing,  or 
equal  things,  are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing  are 
equal. 

"8.  The  whole  is  greater  than  its  part. 
9.  Every  whole  is  equal  to  all  its  parts  taken  to- 
gether. 

A2 


10  GEOMETRY. 

10.  Things  which  coincide,  or  fijl  the  same  space, 
are  identical,  or  mutually  equal  in  all  their  parts. 

11.  All  right  angles  are  equal. 

12.  Angles  that  have  equal  measures,  or  arcs,  are 
equal. 

13.  A  straight  line  is  the  shortest  distance  between 
two  points.  Corollary. — One  side  of  a  triangle  is  less 
than  the  sum  of  the  other  two. 

14.  But  one  straight  line  can  be  drawn  between 
two  points.* 


EXERCISE     WITH     RULE     AND     DIVIDERS     UPON     THE-    RIGHT     LINE     AND 
ANGLE. 

1.  Make  a  line  equal  to  the  sum  of  two  given  lines.     Of  four. 

2.  Make  a  line  equal  to  the  difference  of  two  given  lines. 

3.  Make  a  line  equal  to  five  times  one  given  line  and  six  times 
another. 

4.  Find  how  many  times  one  given  line  is  contained  in  another. 

5.  Find  a  common  measure  of  two  given  lines.t 

6.  Make  a  straight  line  equal  in  length  to  a  broken  line. 

7.  Make  a  straight  line  equal  in  length  approximately  to  a  curve. \ 

8.  With  several  given  points  as  centers,  to  describe  circles  with 
given  lines  as  radii. 

9.  To  find  a  point  which  shall  be  at  given  distances  from  two 
given  points. 

10.  Draw  the  radius  of  a  circle  as  a  chord  of  the  same. 

11.  Make  an  angle  double  a  given  angle.     Triple. 

12.  Measure  the  number  of  degrees  in  a  given  angle  by  means  of  a 
brass  or  paper  circle  or  semicircle,  divided  into  degrees,  called  a  pro- 
tractor. 

13.  Make  an  angle  equal  to  the  sum  of  several  given  angles. 

14.  Draw  a  line  through  a  given  point  parallel  to  a  given  line. 

15.  Draw  through  given  points  several  parallels  to  a  given  line. 

*  A  straight  line  joining  two  points  is  the  direction  of  the  one  from 
the  other.  Two  points  are  said  to  determine  a  line.  Two  points  of 
a  line  being  given,  the  line  is  given ;  for  it  is  the  line  joining  them. 

t  This  is  done  in  a  manner  analogous  to  the  corresponding  opera- 
tion in  Arithmetic  and  Algebra,  by  applying  the  smaller  line  to  the 
larger  as  many  times  as  it  will  go ;  and  the  remainder  to  the  smaller 
given  line,  and  so  on. 

\  This  may  be  done  by  taking  such  small  portions  of  the  curve  as 
ire  nearly  straight. 


THEOREMS.  11 

1G.  Draw  through  a  given  point,  without  a  given  line,  a  line  form- 
ing with  it  a  given  angle. 

17.  To  makfl  ;ai  angle  with  two  given  lines  for  sides. 

18.  In  how  many  points  may  20  lines  cut  each  other,  no  two  of 
which  are  parallel? 

19.  Iu  how  many  when  twelve  of  them  are  parallel  ? 

'20.  In  how  many  when  4  are  parallel  in  one  direction,  5  in  an- 
other, and  6  in  another  ? 

21.  In  how  many  points  will  36  lines  intersect,  24  of  which  pass 
through  the  same  point  ? 


THEOREM    I. 

If  two  triangles  have  two  sides  and  the  included 
angle  in  the  one  equal  to  two  sides  and  the  included 
angle  in  the  other,  the  triangles  will  be  identical,  or 
equal  in  all  respects. 


In  the  two  triangles  ABC,  DEF,  if  the  side 
AB  be  equal  to  the  side  j± 
DE,  the  side  AC  equal 
to  the  side  DF,  and  the 
angle  A  equal  to  the  an- 
gle D,  then  will  the  two 
triangles  be  identical,  or 
equal  in  all  respects. 

For,  conceive  the  triangle  ABC  to  be  applied  to,  or 
placed  on,*  the  triangle  DEF,  in  such  a  manner  that 
the  point  A  may  coincide  with  the  point  D,  and  the 
side  AB  with  the  side  DE,  which  is  equal  to  it. 

Then,  since  the  angle  A  is  equal  to  the  angle  D 
(by  hyp.),f  the  side  AC  must  differ  in  direction  from 
the  side  AB  by  the  same  amount  that  the  side  DF 
does  from  DE  ;  hence  AC  must  take  the  same  direc- 

*  The  student  will  do  well,  at  first,  to  cut  two  triangles  out  of  paste- 
board Of  paper,  and  pUbe  one  upon  the  other;  or  imagine  the  first 
of  (he  above  triangles  to  be  cut  out  of  the  page  and  placed  upon  the 
Other;  or  conceive  the  sides  to  be  fine  wires,  so  that  the  triangle  can 
be  t.iken  oft' the  page. 

t  1 1 yp.  stands  for  hypothesis.  This  term  is  much  used,  and  signifies 
generally  that  what  is  stated  is  given  or  supposed  true  at  the  outset. 


12  METKV. 


tioii  as  DP,  and,  since  AC  is  (by  hyp.)  equal  to  J)F, 
the  point  C  must  fall  on  the  point.  F,  and,  by  Ax.  14, 
the  line  BC  must  fall  on  EF;  thus  the  two  triangles 

coincide  throughout,  and  (Ax.  10)  are  identical. 
Q.  E.  D.* 


THEOREM    II. 

When  two  triangles  have  two  angles  and  the  in- 
cluded side  in  the  one  equal  to  two  angles  and  thr. 
included  side  in  the  other,  the  triangles  are  identical, 
or  have  their  other  sides  and  angles  equal. 

Let  the  two  triangles 
ABC,  DEF  have  the  an- 
gle A  equal  to  the  angle  D, 
the  angle  B  equal  to  the 
angle  E,  and  the  side  AB 
equal  to  the  side  DE  ;  then 
these  two  triangles  will  be 
identical. 

For,  conceive  the  triangle  ABC  to  be  placed  on 
the  triangle  DEF  in  such  manner  that  the  side  AB 
may  fall  exactly  on  the  equal  side  DE.  Then,  since 
the  angle  A  is  equal  to  the  angle  D  (by  hyp.),  the 
side  AC  must  fall  on  the  side  DF ;  and,  in  like  man- 
ner, because  the  angle  B  is  equal  to  the  angle  E,  the 
side  BC  must  fall  on  the  side  EF,  and  the  two  sides 
AC  and  BC  coinciding  respectively  with  the  two 
DE  and  EF,  they  must  meet  in  the  same  point,  that 
is,  the  point  C  must  fall  on  the  point  F.  Thus  the 
three  sides  of  the  triangle  ABC  will  be  exactly  placed 
on  the  three  sides  of  the  triangle  DEF  ;  consequently, 
the  two  triangles  are  identical  (ax.  10),  and  the  other 
two  sides  AC,  BC  will  be  equal  to  the  two  DF,  EF, 
and  the  remaining  angle  C  equal  to  the  remaining 
angle  F.     Q.  E.  D. 

*  These  letters  are  the  initials  of  the  words  "quod  erat  demon- 
strandum," signifying  "which  was  to  be  demonstrated." 


THEOREMS.  13 


THEOREM    HI. 


In  an  isosceles  triangle,  the  angles  at  the  base*  are 
equal.  Or,  if  a  triangle  have,  two  sides  equal,  their 
opposite  angles  will  also  be  equal. 

If  the  triangle  ABC  have  the  side  AB         aa 
equal  to  the  side  AC,  then  will  the  an- 
gle B  be  equal  to  the  angle  C. 

For,  conceive  the  angle  A  to  be  bi-        I 
sected,  or  divided  into  two  equal  parts      / 

by  the  line  AD,  making  the  angle  BAD    / 

equal  to  the  angle  CAD.  B       D       C 

Then  the  two  triangles  BAD,  CAD  have  two 
sides  and  the  contained  angle  of  the  one  equal  to 
two  tides  and  the  contained  angle  of  the  other,  viz., 
the  side  AB  equal  to  AC,  the  angle  BAD  equal  to 
CAD,  and  the  side  AD  common;  therefore  these  two 
triangles  are  identical,  or  equal  in  all  respects  (th.  1)  ; 
and,  consequently,  the  angle  C  equal  to  the  angle  B. 

a  k.d. 

CoroL  1.  Hence  the  line  which  bisects  the  vertical 
angle  of  an  isosceles  triangle  bisects  the  base,  and  is 
also  perpendicular  to  it.     (See  def.  12.) f 

Carol.  2.  Hence,  too,  it  appears  that  every  equi- 
lateral triangle  is  also  equiangular,  or  has  all  its 
angles  equal ;  for  an  equilateral  triangle  is  isosceles, 
whichever  side  may  be  taken  for  the  base. 

*  Th'*  base  of  an  isosceles  triangle  is  the  side  unequal  to  either  of 
ih"  other  two. 

t  The  line  AD  passes  through  the  vertex,  bisects  the  vertical  angle, 

-  the  base,  and  is  perpendicular  to  tin-  base.     Anv  two  of  these 

tour  conditions  determine  the  position  of  the  line,  and  the  other  two 

conditions  follow.      Whence  the  following   theorems   in   addition  to 

cor.  1. 

1.  The  line  ioinxng  the  vertex  of  an  isosceles  triangle  with  the 
middle  of  the  base  bisects  the  vertical  angle,  anil   is  perpendicular 
base. 

be  perpendicular  at  the  middle  of  the  base  passes  through  the 
.  and  bisects  tin-  vertical  angle. 
:?.  The  perpendicular  from  the  vertex  to  the  base  bisects  tin-  base 
and  the  vertical  angle. 

To  prove  each  of  these  independently  will  be  an  exercise. 


14  GEOMETRY. 


THEOREM    IV. 


When  a  triangle  has  two  of  its  angles  equal,  the  sides 
opposite  to  them  are  also  equal. 

If  the  triangle  ABC  have  the  angle 
C  equal  to  the  angle  B,  it  will  also 
have  the  side  AB  equal  to  the  side  AC. 

For  if  not,  let  BA  be  greater  than 
AC,  and  take  BD  equal  to  AC,  and  join 
the  points  C  and  D  by  the  line  CD ; 
then  the  two  triangles  ABC  and  CBD 
having  the  side  BC  common,  the  side   B  c 

BD  of  the  one  equal  (by  construction)  to  the  side  AC 
of  the  other,  and  the  contained  angle  BCA  of  the  for- 
mer equal  to  the  contained  angle  B  of  the  latter,  are 
equal  (by  th.  1) ;  but  the  triangle  CBD  is  evidently 
only  a  part  of  the  triangle  ABC,  and  a  part  can  not 
be  equal  to  the  whole  (ax.  8).  The  hypothesis  must, 
therefore,  be  wrong,  and  AB  can  not  be  greater  than 
AC.  In  a  similar  manner  it  might  be  proved  that 
AC  can  not  be  greater  than  AB ;  hence  AB  =  AC. 
Q.  E.  D.* 

Corol  Hence  every  equiangular  triangle  is  also 
equilateral. 

THEOREM    V. 

When  two  triangles  have  all  the  tnree  sides  in  the 
one  equal  to  all  the  three  sides  in  the  other,  the  trian- 
gles are  identical,  or  have  also  their  three  angles  equal, 
each  to  each. 

Let  the  two  triangles  ABC,  ABD  0 

have  their  three  sides  respectively         ^^^T\ 

equal,  viz.,  the  side  AB  equal  to  AB,  A<^- * — -^B 

AC  to  AD,  and  BC  to  BD ;  then     ^"^\j/ 
shall  the  two  triangles  be  identical,  rj 

*  The  kind  of  demonstration  employed  here  is  called  negative  rea 
soning,  because,  by  proving  that  the  negative  can  not  be  true,  it  proves 
the  affirmative.  It  is  also  called  the  reductio  ad  absnrdum,  because 
it  proves  that  the  denial  of  the  proposition  leads  to  an  absurdity. 


THEOREMS.  15 

or  have  their  angles  equal,  viz.,  those  angles  that  are 
opposite  to  the  equal  sides ;  that  is  to  say,  the  angle 
BAC  to  the  angle  BAD,  the  angle  ABC  to  the  angle 
ABD,  and  the  angle  C  to  the  angle  D. 

For,  conceive  the  two  triangles  to  be  joined  togeth- 
er by  their  longest  equal  sides,  and  draw  the  line  CD. 

Then,  in  the  triangle  ACD,  because  the  side  AC  is 
(by  hyp.)  equal  to  AD,  the  angle  ACD  is  equal  to  the 
angle  ADC  (th.  3).  In  like  manner,  in  the  triangle 
BCD,  the  angle  BCD  is  equal  to  the  angle  BDC,  be- 
cause the  side  BC  is  equal  to  BD.  Hence,  then,  the 
angle  ACD  being  equal  to  the  angle  ADC,  and  the 
angle  BCD  to  the  angle  BDC,  by  equal  additions  the 
sum  of  the  two  angles  ACD,  BCD,  is  equal  to  the 
sum  of  the  two  ADC,  BDC  (ax.  2),  that  is,  the  whole 
angle  ACB  equal  to  the  whole  angle  ADB. 

Since,  then,  the  two  sides  AC,  CB  are  equal  to 
the  two  sides  AD,  DB,  each  to  each  (by  hyp.),  and 
their  contained  angles  ACB,  ADB,  also  equal,  the 
two  triangles  ABC,  ABD  are  identical  (th.  1),  and 
have  the  other  angles  equal,  viz.,  the  angle  BAC  to 
the  angle  BAD,  and  the  angle  ABC  to  the  angle 
ABD.     Q.  E.  D. 

General  Scholium.  From  the  foregoing  propositions 
it  appears  that  triangles  will  be  identical  when  they 
have  two  sides  and  the  included  angle,  two  angles 
and  the  included  side,  or  three  sides  of  the  one  equal 
to  the  same  in  the  other.* 

EXERCISES. 

1.  To  make  a  triangle  when  two  sides  and  the  angle  which  they 
form  are  given. 

*  A  triangle  is  composed  of  six  elements,  three  sides  and  three 
angles.  It  is  only  necessary,  in  general,  that  three  of  these,  provided 
one  be  a  side,  in  one  triangle  should  be  equal  to  the  same  in  the 
other  to  render  the  triangles  equal.  (See  cor.  7,  th.  15,  and  see  prob. 
8  for  an  exception.) 

Three  given  j>arts  are  also  said  to  determine  a  triangle;  by  which 
is  to  l)i-  understood,  that  with  the  three  given  parts  only  one  triangle 
can  he  formed,  or  that  all  which  may  be  formed  with  them  will  be 
identical.  This  remark,  like  the  above,  does  not  include  the  case 
where  three  angles  are  given. 


16  GEOMETRY. 

2.  When  two  angles  and  the  side  of  the  triangle  between  them  are 
given. 

3.  On  a  given  line  to  construct  an  equilateral  triangle. 

4.  To  construct  an  isosceles  triangle  with  a  given  base  and  given 
side. 

5.  To  construct  a  triangle  with  three  given  lines  for  sides. 

N.B.  The  given  sides  must  be  subject  to  the  condition  expressed 
in  the  corollary  to  ax.  13. 

THEOREM    VI. 

When  one  line  meets  another,  the  angles  which  the 
first  line  makes  on  the  same  side  of  the  second  are 
together  equal  to  two  right  angles. 

Let  the  line  AB  meet  the  line  CD  ; 
then  will  the  two  angles  ABC,  ABD, 
taken  together,  be  equal  to  two  right 
angles. 

For,  suppose  BE  drawn  perpen-  

dicular  to  CD.    Then  the  two  angles  c  B 

CBA,  ABD  fill  the  same  angular  space  with  the  two 
CBE,  EBD ;  but  the  latter  are  right  angles,  hence 
the  former  are  together  equal  to  two  right  angles. 

Corol.  1.  Conversely,  if  the  two  angles  ABC,  ABD, 
on  opposite  sides  of  the  line  AB,  make  up  together 
two  right  angles,  then  CB  and  BD  form  one  contin- 
ued right  line  CD.  For  no  other  line  from  the  point 
B  than  BD,  the  continuation  of  CB,  can  form  with  BA 
an  angle  equal  to  ABD,  which,  added  to  ABC,  makes 
two  right  angles,  by  the  above  theorem. 

Corol.  2.  All  the  angles  which  can  be  made  at  any 
point  B,  by  any  number  of  lines,  on  the  same  side  of 
the  right  line  CD,  are,  when  taken  all  together,  equal 
to  two  right  angles,  since  they  fill  the  same  angular 
space. 

Corol.  3.  And  as  all  the  angles  that  can  be  made 
on  the  other  side  of  the  line  CD  are  also  together 
equal  to  two  right  angles  ;  therefore,  all  the  angles 
that  can  be  made  quite  round  a  point  B,  by  any  num- 
ber of  lines,  are  together  equal  to  four  right  angles. 

Corol.  4.  Hence,  also,  the  whole  circumference  of 


THEOREMS.  17 

a  circle,  being  the  sum  of  the  measures  of  all 

Che  angles  that  can  be  made  about  the  center 

P,  is  the  measure  of  lour   right  angles ;   a 

semicircle,  or  180  degrees,  is  the  measure  ol 

two  right  angles;  and  a  quadrant,  or  90  degrees,  the 

measure  of  one  right  angle.     (See  def.  49.) 

Two  anglei  which  together  amount  to  a  right  angle,  or  90°,  are 
called  complement*  of  each  other.  Thus,  40°  is  the  complement  of 
.a. I  50°  of  40°. 

Two  angle*  which  together  ecpial  two  right  angles,  or  180°,  as" 
111)    ami  7U-\  are  called  supplements  of  each  other. 

THEOREM    VII. 

}\/ien  two  lines  intersect  each  other,  the  opposite 
angles  are  equal. 

Let  the  two  lines  AB,  CD  inter- 
sect in  the  point  E  ;  then  will  the 
angle  AEC  be  equal  to  the  angle 
BED,  and  the  angle  AED  be  equal  A 
to  the  angle  CEB. 

For  EA  is  exactly  the  opposite  -° 

direct  ion  from  EB,  and  ED  exactly  the  opposite  di- 
m-lion from  EC.  Hence  their  difference  of  direction 
will  be  the  same.     Q.  E.  D.     (See  def.  10.) 

THEOREM    VIII. 

^Yhen  one  side  of  a  triangle  is  produced,  the  out- 
toard  angle  is  greater  than  either  of  the  two  inward 
opposite  angles. 

Let  ABC  be  a  triangle,  having 
the  side  AB  produced  to  D;  then 
will  the  outward  angle  CBD  be 
greater  than  either  of  the  inward 
opposite  angles  A  or  C. 

For,  conceive  the  side  BC  to  be 
v<]  in  the  point  E,  and  draw 
the  l.ne  AE,  producing  it  till  EF  be 
equal  to  AE;  and  join  BF. 


18  GEOMETRY. 

Then,  since  the  two  triangles  AEC,  BEF  have  the 
side  AE  =s  the  side  EF,  and  the  side  CE  =  the  side 
BE  (by  constr.),  and  the  included  or  opposite  angles 
at  E  also  equal  (th.  7),  therefore  those  two  triangles 
are  equal  in  all  respects  (th.  1),  and  have  the  angle 
C  =  the  corresponding  angle  EBF.  But  the  angle 
CBD  is  greater  than  the  angle  EBF  (ax.  8) ;  conse- 
quently, the  said  outward  angle  CBD  is  also  greater 
than  the  angle  C. 

In  like  manner,  if  CB  be  produced  to  G,  and  AB 
be  bisected,  it  may  be  shown  that  the  outward  angle 
ABG,  or  its  equal  (th.  7)  CBD,  is  greater  than  the 
other  inward  angle  A. 

THEOREM    IX. 

The  greater  side  of  every  triangle  is  opposite  to  the 
greater  angle,  and  the  greater  angle  opposite  to  the 
greater  side. 

Let  ABC  be  a  triangle,  having  the 
side  AB  greater  than  the  side  BC ;  then 
will  the  angle  ACB,  opposite  the  greater 
side  AB,  be  greater  than  the  angle  A, 
opposite  the  less  side  CB. 

For,  on  the  greater  side  ABD  take  the 
part  BD,  equal  to  the  less  side  BC,  and  c  B 

join  CD.  Then,  since  ACD  is  a  triangle,  the  out- 
ward angle  BDC  is  greater  than  the  inward  opposite 
angle  A  (th.  8).  But  the  angle  BCD  is  equal  to  the 
said  outward  angle  BDC,  because  the  triangle  BDC 
is  isosceles,  BD  being  equal  to  BC  (th.  3).  Conse- 
quently, the  angle  BCD,  also,  is  greater  than  the  an- 
gle A.  And,  since  the  angle  BCD  is  only  a  part  of 
ACB,  much  more  must  the  whole  angle  ACB  be 
greater  than  the  angle  A.     Q.  E.  D. 

Again,  conversely,  in  the  given  triangle  ABC,  if 
the  angle  C  be  greater  than  the  angle  A,  then  will 
the  side  AB,  opposite  the  former,  be  greater  than  the 
side  BC,  opposite  the  latter. 

For  if  AB  be  not  greater  than  BC,  it  must  be 


THEOREMS.  19 

either  equal  to  it  or  less  than  it.  But  it  can  not 
be  equal,  for  then  the  angle  C  would  be  equal  to 
the  angle  A  (th.  3),  which  it  is  not,  by  the  sup- 
position. Neither  can  it  be  less,  for  then  the  angle 
C  would  be  less  than  the  angle  A,  by  the  former  part 
of  this  theorem,  which  is  also  contrary  to  the  sup- 
position. The  side  AB,  then,  being  neither  equal 
to  BC,  nor  less  than  it,  must  necessarily  be  greater. 
Q.  E.  D. 

THEOREM    X. 

When  a  line  intersects  two  parallel  lines  obliquely  it 
will  form  with  them  four  acute  angles  and  four  obtuse  ; 
the  four  acute  will  be  equal  to  one  another,  and  the  four 
obtuse. 

This  follows  from  defini- 
tions 8  and  10,  and  th.  7. 

Corol.  1.  Any  one  of  the 
acute  and  any  one  of  the  ob- 
tuse will  be  supplements  of 
each  other.     (See  th.  6.) 

Corol.  2.  If  one  of  the  an- 
gles be  right,  the  whole  eight 
will  be  right  angles. 

Scholium  1.  The  line  cutting  the  two  parallels  is 
sometimes  called  the  secant  line.  The  two  angles 
within  the  parallels,  and  on  different  sides  of  the  se- 
cant line,  are  commonly  called  alternate  internal  or 
interior  angles,  or  simply  alternate  angles.  The  two 
outside  the  parallels,  and  on  different  sides  the  secant, 
alternate  external  angles ;  and  the  two  on  the  same, 
side  of  the  secant,  one  within  and  the  other  without 
the  parallels,  and  not  adjacent,  are  called  opposite 
internal  and  external,  or  outward  and  inward  on  the 
tame  side. 

Scholium  2.  The  distance  of  two  parallel  lines  is  the 
length  of  the  line  between  them,  drawn  perpendicular 
to  both,  as  FH  or  EG  in  the  next  diagram. 


20  GEOMETRY. 


THEOREM    XI. 

When  one  straight  line  meets  two  others  so  as  to 
make  equal  angles  with  them,  the  latter  are  parallel ;  in 
other  words,  if  two  lines  make  the  same  angle  with  a 
third,  they  will  be  parallel  to  each  other. 

For,  having  the  same  difference  of  direction  from 
the  same  line,  they  must  have  the  same  direction 
with  one  another,  and  are  therefore  (def.  8)  parallel. 

Note. — Two  lines  parallel  to  a  third  are  parallel  to 
one  another.     This  follows  from  def.  8. 

THEOREM    XII. 

Parallel  lines  are  every  where  equally  distant. 
To  prove  this  it  will  only  h  g- 


be  necessary  to  prove  any  c 
two     perpendiculars,     FH 
and  EG,  drawn  at  random 


between  them,  equal.    Join  A      F  E    B 

EH.  Then  the  lines  FH  and  EG  both  being  at  right 
angles  to  AB,  will,  by  the  last  theorem,  be  parallel ; 
and  the  two  triangles  EFH,  EGH  will  have  the  side 
EH  common,  and  the  two  angles  adjacent  this  side 
in  the  one  equal  to  the  same  in  the  other  ;  hence  (th. 
2)  these  triangles  are  equal  in  all  respects,  and,  there- 
fore, the  side  FH  of  the  one  equal  to  the  correspond- 
ing side  EG  of  the  other. 

Carol.  1.  Parallel  lines,  being  every  where  at  the 
same  distance,  however  far  produced,  can  never  meet. 
This  is  sometimes  expressed  by  saying  that  they 
meet  at  an  infinite  distance. 

Corol.  2.  If  the  extremities  of  twTo  equal  perpen- 
diculars to  a  given  line  be  joined,  a  parallel  will  be 
obtained. 


THEOREMS.  21 


THEOREM    Mil. 

117/en  one  side  of  a  triangle  is  produced,  the  out- 
ward  angle  is  equal  to  both  the  inward  opposite  angles 
taken  together.' 

Let  the  side  BC  of  the 
triangle  ABC  be  produced 
t<>  I) ;  then  will  the  out- 
ward angle  ACD  be  equal 
to  the  sum  of  the  two  in- 
ward opposite  angles  A 
and  B. 

For,  conceive  CE  to  be  drawn  parallel  to  the  side 
IlB  of  the  triangle.  Then  AC,  meeting  the  two  par- 
allels BA,  CE,  makes  the  alternate  angles  A  and  ACE 
■qua!  (th.  10).  And  BD,  cutting  the  same  two  par- 
allels BA,  CE,  makes  the  inward  and  outward  angles 
on  the  same  side,  B  and  ECD,  equal  to  each  other 
(th.  10).  Therefore,  by  equal  additions,  the  sum  of 
tin.'  two  angles  A  and  B  is  equal  to  the  sum  of  the 
two  ACE  and  ECD,  that  is,  to  the  whole  angle  ACD 
(by  ax.  2).     Q.E.D. 

THEOREM    XIV. 

If  two  angles  stand  upon  the  same  base,  the  vertex 
of  one  being  within  that  of  the  other,  the  latter  will  be 
the  less  angle. 

In  the  diagram  .the  angle  ABC  is 
less  than  the  angle  ADC  ;  for  (th.  8) 
ADE  >  ABD,  and  EDC  >  EBC  .-. 
by  equal  addition  ADE  +  EDC  > 
ADD  +  EBC,  or  ADC  >  ABC. 
Q.  E.  D.  a- 

*  The  two  opposite  angles  are  the  two  neither  of  which  is  adjacent 
the  outward  angle. 

d  to  1>»'  adjacent  to  another  when  the  two  have  a 
coiinii'iii  vertex  and  a  common  side.  An  angle  is  said  to  be  adjacent 
to  a  line  when  that  line  is  one  of  its  sides. 

N.B. — It  will  be  found  convenient,  when  angles  have  been  proved 
equal,  to  mark  them  with  the  same  number  of  dots  as  in  the  figure. 


22  GEOMETRY. 


THEOREM    XV. 

In  any  triangle  the  sum  of  all  the  three  an± 
equal  to  two  right  angles. 

Let  ABC  be  any  plane  triangle  ; 
then  the  sum  of  the  three  angles  A  -f- 
B  +  C  is  equal  to  two  right  angles. 

For,  let  the  side  AB  be  produced 
to  D.  Then  the  outward  angle  CBD  A 
is  equal  to  the  sum  of  the  two  inward  opposite  angles 
A  +  C  (th.  13).  To  each  of  these  equals  add  the  in- 
ward angle  B;  then  will  the  sum  of  the  three  inward 
angles  A  +  B  +  C  be  equal  to  the  sum  of  the  two 
adjacent  angles  ABC  -f-  CBD  (ax.  2).  But  the  sum 
of  these  two  last  adjacent  angles  is  equal  to  two  right 
angles  (th.  6).  Therefore,  also,  the  sum  of  the  three 
angles  of  the  triangle  A  +  B  -f  C  is  equal  to  two 
right  angles  (ax.  1).     Q.  E.  D. 

Corol.  1.  If  two  angles  in  one  triangle  be  equal  to 
two  angles  in  another  triangle,  the  third  angles  will 
also  be  equal;  for  they  make  up  two  right  angles  in 
both. 

Corol.  2.  Two  right-angled  triangles  will  be  equi- 
angular when  they  have  an  acute  angle  in  each  equal. 

Corol.  3.  The  sum  of  two  angles  of  any  triangle 
and  the  third  angle  are  supplements  of  each  other. 

Corol.  4.  If  one  angle  in  one  triangle  be  equal  to 
one  angle  in  another,  the  sums  of  the  remaining  an- 
gles will  also  be  equal  (ax.  3). 

Corol.  5.  If  one  angle  of  a  triangle  be  right,  the 
sum  of  the  other  two  will  also  be  equal  to  a  right 
angle,  and  each  of  them  singly  will  be  acute,  or  less 
than  a  right  angle,  and  wall  be  the  complement  of  the 
other. 

Corol.  6.  The  two  least  angles  of  every  triangle 
are  acute,  or  each  less  than  a  right,  angle.  In  other 
words,  there  can  be  but  one  right  angle,  or  one  obtuse 
angle  in  a  triangle. 

Corol.  7.  Any  two  angles  and  a  side  of  one  trian- 


THEOREMS.  23 

gle  being  equal  to  the  same  in  another,  the  triangles 
will  be  equal  (th.  2). 

Carol.  8.  One  side  and  an  acute  angle  of  a  right- 
angled  triangle  being  equal  to  the  same  in  another, 
the  triangles  are  equal. 


THEOREM    XVL 

The  sum  of  all  the  inward  angles  of  a  polygon  is 
equal  to  twice  as  many  right  angles,  vmnting  four,  as 
the  figure  has  sides. 

Let   ABCDE   be   any   figure  ;  D 

then  the  sum  of  all  its  inward  an-            /T\ 
gles,   A-J-B  +  C  +  D  +  E,   is        /       N. 
equal  to  twice  as  many  right  an-  E^-- jF  _.____^c 

gles,  wanting  four,  as  the  figure      \        /\         / 
has  sides.  \    /     \     / 

For,  from  anv  point  F,  within  V Jy 

it,  draw  lines,  FA,  FB,  FC,  &c,  A  B 

to  all  the  angles,  dividing  the  polygon  into  as  many 
triangles  as  it  has  sides.  Now  the  sum  of  the  three 
angles  of  each  of  these  triangles  is  equal  to  two  right 
angles  (th.  15)  ;  therefore,  the  sum  of  the  angles  of 
all  the  triangles  is  equal  to  twice  as  many  right 
angles  as  the  figure  has  sides.  But  the  sum  of  all 
the  angles  about  the  point  F,  which  are  so  many  of 
the  angles  of  the  triangles,  but  no  part  of  the  inward 
angles  of  the  polygon,  is  equal  to  four  right  angles 
(corol.  3,  th.  6),  and  must  be  deducted  out  of  the  for- 
mer sum.  Hence  it  follows  that  the  sum  of  all  the 
inward  angles  of  the  polygon  alone,  A  +B  +  C  +  D 
+  E,  is  equal  to  twice  as  many  right  angles  as  the 
figure  has  sides,  wanting  the  said  four  right  angles. 
QE.D. 

Corol.  1.  In  any  quadrangle,  the  sum  of  all  the 
four  inward  angles  is  equal  to  four  right  angles. 

Corol.  2.  Hence,  if  three  of  the  angles  be  right 
ones,  the  fourth  will  also  be  a  right  angle. 

Carol.  3.  And  if  the  sum  of  two  of  the  four  angles 


24  GEOMETRY. 

be  equal  to  two  right  angles,  the  sum  of  the  remain- 
ing two  will  also  be  equal  to  two  right  angles. 

Corol.  4.  The  sum  of  the  angles  of  a  pentagon  is 
5X2  —  4  =  6  right  angles.  Of  a  hexagon,  6X2  — 
4  =  8  right  angles.*  Of  a  polygon  of  n  sides  (n  x  2  — 
4)   right  angles. 

The  rule  may  be  thus  expressed :  to  obtain  the 
sum  of  the  angles  of  a  polygon,  double  the  number 
of  sides  and  subtract  4,  the  right  angle  being  the 
unit  of  measure-! 

THEOREM    XVII. 

A  perpendicular  is  the  shortest  line  that  can  be 
drawn  from  a  given  point  to  an  indefinite  line.  And, 
of  any  other  lines  drawn  from  the  same  point,  those 
that  are  equally  distant  from  the  perpendicular  are 
equal,  and  those  nearest  the  perpendicular  are  less  than 
those  more  remote. 

If  AB,  AC,  AD,  &c,  be  lines 
drawn  from  the  given  point  A,  to 
the  indefinite  line  DE,  of  which 

AB  is  perpendicular ;  then  shall 

the  perpendicular  AB  be  less  than  C  B 

AC,AE  =  AC  if  BE  =  BC,and  AC<AD  if  BC<BD. 

For,  the  angle  B  being  a  right  one,  the  angle  C  of 

*  Each  angle  of  a  regular  hexagon  would  be  the  sixth  part  of  8 
right  angles,  or  £  or  -J  or  li  right  angles,  i.  e.,  120°,  90° 
being  a  right  angle,  this  is  i  of  360°,  or  4  right  angles ; 
so  that  if  three  regular  hexagons  were  placed  together 
they  would  fill  up  the  whole  angular  space  about  a 
point. 

This  would   not  be   the  case  with   pentagons,  for 

5X2 4 

^ =  | '■  =  Vs,  which,  multiplied  by  3,  gives  3|  <  4  right  an- 
gles, and,  multiplied  by  4,  gives  4|  >  4  right  angles,  so  that  they  would 
fit  together  neither  by  threes  nor  fours. 

It  will  be  found,  on  examination,  that  no  other  figures  except  squares 
and  triangles  have  this  same  property  with  the  hexagon.  Hence 
these  three  kinds  of  figures  alone  are  employed  for  paving  blocks. 

t  The  above  applies  only  to  convex  polygons,  that  is,  those  in  which 
the  angles  point  outward.  No  convex  polygon  can  have  more  than 
three  acute  angles. 


THEOREMS. 


25 


the  triangle  ABC  is  acute  (by  cor.  6,  th.  15),  and 
therefore  less  than  the  angle  B.  But  the  less  angle 
of  a  triangle  is  subtended  by  the  less  side  (th.  9). 
Therefore  the  side  AB  is  less  than  the  side  AC. 

Again,  in  the  right-angled  triangles  ABC,  ABE  the 
two  sides  AB,  BC  being  respectively  equal  to  the  two 
AB,  BE,  the  third  sides  are  equal  (th.  J). 

Carol.  Every  point,  as  A,  of  a  perpendicular  at 
the  middle  of  a  given  line  CE  is  equally  distant  from 
its  extremities  C  and  E. 

Finally,  the  angle  ACB  being  acute,  or  less  than  a 
right  angle,  as  before,  the  adjacent  angle  ACD  will 
be  greater  than  a  right  angle,  or  obtuse  (by  th.  6)  ; 
consequently,  the  angle  D  is  acute  (corol.  6,  th.  15), 
and  therefore  is  less  than  the  angle  C.  And  since 
the  less  side  is  opposite  to  the  less  angle,  therefore 
the  side  AC  is  less  than  the  side  AD.     Q.  E.  D. 

Corol.  The  least  distance  of  a  given  point  from  a 
line  is  the  perpendicular.  For  if  it  were  an  oblique 
line  the  perpendicular  would  be  shorter,  and  thus  less 
than  the  least  distance,  which  is  impossible. 

THEOREM    XVIIf. 


Every  point  out  of  a  perpendicular  at  the  middle  of 
a  given  line  is  at  unequal  distances  from  the  extremities 
of  the  line. 

Let  DC  be  a  perpendicular  at  the 
middle  of  AB,  and  I  a  point  out  of  the 
perpendicular,  then  shall  IB  <  IA. 
For  join  BD ;  then,  BI  <  BD  +  Dl  ; 
or  since  BD  =  AD  (th.  17),  BI  <  AD  . 
+  DI,  or  BI  <  AI.     Q.  E.  D. 

Scholium.  There  can  be  but  one 
perpendicular  through  a  given  point 
to  a  given  line.  For  there  can  be 
but  one  line  through  the  same  point  in  the  same 
direction,  or  having  the  same  difference  of  direction 
from  a  given  line. 

B 


26  GEOMETRY. 


THEOREM    XIX. 

The  opposite  sides  and  angles  of  a  parallelogram 
are  equal  to  each  other,  and  the  diagonal  divides  it 
into  two  equal  triangles. 

Let  ABDC  be  a  parallelogram,  of  ^  B 

which  the  diagonal  is  BC  ;  then  will    r *?y 

its  opposite  sides  and  angles  be  equal     \        /''     \ 

to  each  other,  and  the  diagonal  BC       XyS \ 

will  divide  it  into  two  equal  parts,       C  D 

or  triangles. 

For,  since  the  sides  AB  and  DC  are  parallel,  as 
also  the  sides  AC  and  BD  (def.  30),  and  the  line  BC 
meets  them  ;  therefore  the  alternate  angles  are  equal 
(th.  10),  namely,  the  angle  ABC  to  the  angle  BCD, 
and  the  angle  ACB  to  the  angle  CBD.  Hence  the 
two  triangles,  having  two  angles  in  the  one  equal  to 
two  angles  in  the  other,  have  also  their  third  angles 
equal  (cor.  1,  th.  15),  namely,  the  angle  A  equal  to 
the  angle  D,  which  are  two  of  the  opposite  angles  of 
the  parallelogram. 

Also,  if  to  the  equal  angles  ABC,  BCD  be  added 
the  equal  angles  CBD,  ACB,  the  wholes  will  be  equal 
(ax.  2),  namely,  the  whole  angle  ABD  to  the  whole 
ACD,  which  are  the  other  two  opposite  angles  of  the 
parallelogram.     Q  E.  D. 

Again,  since  the  two  triangles  are  mutually  equi- 
angular, and  have  a  side  in  each  equal,  viz.,  the  com- 
mon side  BC  ;  therefore  the  two  triangles  are  identi- 
cal (th.  2),  or  equal  in  all  respects,  namely,  the  side 
AB  equal  to  the  opposite  side  DC,  and  x\C  equal  to 
the  opposite  side  BD,*  and  the  whole  triangle  ABC 
equal  to  the  whole  triangle  BCD.     Q.  E.  D. 

Corol.  1.  Hence,  if  one  angle  of  a  parallelogram 
be  a  right  angle,  all  the  other  three  will  also  be  right 

*  The  student  will  observe  that  in  identical  triangles  the  equal 
sides  are  opposite  equal  angles.  Thus,  in  the  diagram,  the  side  BD 
opposite  the  angle  C,  in  the  lower  triangle,  is  equal  to  the  side  AC, 
opposite  the  equal  angle  B  in  the  upper. 


THEOREMS.  27 

angles,  and  the  parallelogram  a  rectangle.  (See 
cors.  to  th.  16.) 

Corol.  2.  Hence,  also,  the  sum  of  any  two  adjacent 
angles  of  a  parallelogram  is  equal  to  two  right  angles. 

Carol.  3.  If  two  parallelograms  have  an  angle  in 
each  equal,  the  parallelograms  are  equiangular. 

THEOREM    XX. 

Every  quadrilateral  whose  opposite  sides  are  equal 
is  a  parallelogram,  or  has  its  opposite  sides  parallel. 

Let  ABDC  be  a  quadrangle  hav-    .  B 

ing  the  opposite  sides  equal,  namely, 
the  side  AC  equal  to  BD,  and  AB 
equal  to  CD  ;  then  shall  these  equal 
sides  be  also  parallel,  and  the  figure 
a  parallelogram. 

For,  let  the  diagonal  BC  be  drawn.  Then  the 
triangles  ABC,  CBD  being  mutually  equilateral  (by 
hyp.),  they  are  also  mutually  equiangular  (th.  5),  or 
have  their  corresponding  angles  equal ;  consequently, 
the  opposite  sides,  having  the  same  difference  of 
direction  in  opposite  ways  from  the  same  line  BC, 
have  the  same  direction  one  way,  and  are  parallel 
(def.  8) ;  viz.,  the  side  AB  parallel  to  DC,  and  AC 
parallel  to  BD,  and  the  figure  is  a  parallelogram  (def. 
30).     Q.  E.  D. 

THEOREM    XXI. 

Those  lines  which  join  the  corresponding  extremities 
of  two  equal  and  parallel  lines  are  themselves  equal 
and  parallel. 

Let  AB,  DC  be  two  equal  and  parallel  lines ;  then 
will  the  lines  AC,  BD,  which  join  their  extremes,  be 
also  equal  and  parallel.     [See  the  fig.  above.] 

For,  draw  the  diagonal  BC.  Then,  because  AB 
and  DC  are  parallel  (by  hyp.),  the  angle  ABC  is 
equal  to  the  alternate  angle  DCB  (th.  10).  Hence, 
then,  t ho  two  triangles  having  two  sides  and  the  con- 


28  GEOMETRY. 

tained  angles  equal,  viz.,  the  side  AB  equal  to  the 
side  DC,  and  the  side  BC  common,  and  the  contained 
angle  ABC  equal  to  the  contained  angle  DCB,  they 
have  the  remaining  sides  and  angles  also  respectively 
equal  (th.  1)  ;  consequently  AC  is  equal  to  BD,  and 
also  parallel  to  it  (th.  1 1).     Q.  E.  D. 

General  Scholium.  From  the  foregoing  theorems  it 
appears  that  a  quadrilateral  will  be  a  parallelogram : 
1.  When  it  has  its  opposite  sides  parallel  ;  2.  When  it 
has  its  opposite  sides  equal ;  3.  When  it  has  two  of  its 
opposite  sides  equal  and  parallel.  A  quadrilateral  is 
also  a  parallelogram  :  4.  When  two  sides  are  parallel 
and  two  opposite  angles  equal  ;  5.  When  the  opposite 
angles  are  equal.  The  proof  of  the  last  two  cases  is 
left  as  an  exercise  for  the  learner. 


THEOREM    XXII. 

Parallelograms,  as  also  triangles,,  standing  on  the 
same  base,  and  between  the  same  parallels,  are  equal  to 
each  other. 

Let   ABCD,  d 
ABEF  be  two 

parallelograms, 

and  ABC,  ABF 

two     triangles, 

standing  on  the 

same  base  AB,  and  between  the  same  parallels  AB, 

DE  ;  then  will  the  parallelogram  ABCD  be  equal  to 

the  parallelogram  ABEF,  and  the  triangle  ABC  equal 

to  the  triangle  ABF. 

For  the  two  triangles  ADF,  BCE  are  equiangular, 
having  their  corresponding  sides  in  the  same  direc- 
tion ;  and  having  the  two  corresponding  sides  AD, 
BC  equal  (th.  19),  being  opposite  sides  of  a  parallel- 
ogram, these  two  triangles  are  identical,  or  equal  in 
all  respects  (th.  2).  If  each  of  these  equal  triangles, 
then,  be  taken  from  the  whole  space  ABED,  there 
will  remain  the  parallelogram  ABEF  in  the  one  case, 


THEOREMS.  29 

equal  to  the  parallelogram  ABCD  in  the  other  (by 
ax.  3). 

Also,  the  triangles  ABC,  ABF,  on  the  same  base 
AB,  and  between  the  same  parallels,  are  equal,  being 
the  halves  of  the  said  equal  parallelograms  (th.  11)).* 
Q.  E.  D. 

Carol.  1.  Parallelograms,  or  triangles,  having  .the 
same  base  and  altitude  are  equal.  For  the  altitude 
is  the  same  as  the  perpendicular  or  distance  between 
the  two  parallels,  which  is  every  where  equal,  by  the- 
orem 12. 

Corol.  2.  Parallelograms,  or  triangles,  having  equal 
bases  and  altitudes  are  equal.  For,  if  the  one  figure 
be  applied  with  its  base  on  the  other,  the  bases  will 
coincide,  or  be  the  same,  because  they  are  equal ;  and 
so  the  two  figures,  having  the  same  base  and  altitude, 
are  equal. 

THEOREM    XXIII. 

If  a  parallelogram  and  a  trian gle' stand  on  the  same 
base,  and  between  the  same  parallels,  the  parallelogram 
will  be  double  the  triangle,  or  the  triangle  half  the  par- 
allelogram. 

Let  ABCD  be  a  parallelogram,  and  p_ 
ABE  a  triangle,  on  the  same  base  AB, 
and  between  the  same  parallels  AB,  DE  ; 
then  will  the  parallelogram  ABCD  be 
double  the  triangle  ABE,  or  the  triangle 
half  the  parallelogram. 

For,  draw  the  diagonal  AC  of  the  par- 
allelogram, dividing  it  into  two  equal  parts  (th.  19). 
Then,  because  the  triangles  ABC,  ABE  on  the  same 
base,  and  between  the  same  parallels,  are  equal  (th. 
22) ;  and  because  the  one  triangle  ABC  is  half  the 
parallelogram  ABCD  (th.  19),  the  other  equal  triangle 

*  The  trianglei  being  raven,  with  their  vertices  C  and  F  taken  at 
pleasure  in  the  line  DE,  the  lines  BE  and  AD  must  be  drawn  parallel 
t<>  tin-  tides  At'.  BC  of  the  triangles,  to  complete  tin-  parallelograms 

The  above  theorem  may  be  proved  by  th.  1,  and  alio  by  th.  5. 


30  GEOMETRY. 

ABE  is  also  equal  to  half  the  same  parallelogram 
ABCD.     Q.  E.  D. 

Corol.  A  triangle  is  equal  to  half  a  parallelogram 
of  the  same  base  and  altitude. 

THEOREM    XXIV. 

The  complements  of  the  parallelograms  which  are 
about  the  diagonal  of  any  parallelogram  are  equal  to 
each  other. 

Let  AC  be  a  parallelogram,  BD  a     d     G 
diagonal,  EIF   parallel   to  AB   and       f5T 
DC,  and   GIH    parallel  to  AD  and       ' 
BC,  making  AI,  IC  complements  to  E 
the  parallelograms  EG,  HF,  which     l      l- 
are  about  the  diagonal  DB  ;  then  will 
the  complement  AI  be  equal  to  the  complement  IC. 

For,  since  the  diagonal  DB  bisects  the  three  par- 
allelograms AC,  EG,  HF  (th.  19)  ;  therefore,  the 
whole  triangle  DAB  being  equal  to  the  whole  tri- 
angle DCB,  and  the  parts  DEI,  IHB  respectively 
equal  to  the  parts  DGI,  IFB,  the  remaining  parts 
AI,  IC  must  also  be  equal  (by  ax.  3).     Q.  E.  D. 

THEOREM    XXV. 

A  trapezoid  is  equal  to  half  a  parallelogram,  whose 
base  is  the  sum  of  the  two  parallel  sides,  and  its  alti- 
tude the  perpendicular  distance  between  them. 

Let  ABCD  be  the  trapezoid,  having  its  ®_ 
two  sides  AB,  DC  parallel ;  and  in  AB 
produced  take  BE  equal  to  DC,  so  that 
AE  may  be  the  sum  of  the  two  parallel 
sides  ;  produce  DC  also,  and  let  EF,  GC,  A 
BH  be  all  three  parallel  to  AD.  Then  is  AF  a  par- 
allelogram of  the  same  altitude  with  the  trapezoid 
ABCD,  having  its  base  AE  equal  to  the  sum  of  the 
parallel  sides  of  the  trapezoid  ;  and  it  is  to  be  proved 
that  the  trapezoid  ABCD  is  equal  to  half  the  parallel- 
ogram AF. 


( 

Now,  since  triangles,  or  parallelograms,  o/  equal 
bases  and  altitudes,  are  equal  (corol.  2,  ih.  22),  the 
parallelogram  DG  is  equal  to  the  parallelogram  HE, 
and  the  triangle  CGB  equal  to  the  triangle  CHB ; 
consequently,  the  line  BC  bisects  or  equally  divides 
the  parallelogram  AF,  and  ABCD  is  the  half  of  it. 
Q.  E.  D. 


THEOREM    XXVI. 


In  any  right-angled  triangle,  the  square  of  the  hy- 
pothenuse  is  equal  to  the  sum  of  the  squares  of  the 
other  two  sides.  , 

Let  ABC  be  a  right-angled 
triangle,  having  the  right  angle 
A  ;  then  will  the  square  of  the 
hypothenuse  BC  be  equal  to  the 
sum  of  the  squares  of  the  other 
two  sides  AC,  AB.  Or  BC2  = 
AC2  +  AB3. 

For,  on  BC  describe  the 
square  BE,  and  on  AC,  AB,  the 
squares  CH,  BG ;  then  draw  AL 
parallel  to  BD,  and  join  CF,  AD. 

Now,  because  the  line  AB  meets  the  two  AG,  AC, 
so  as  to  make  the  sum  of  the  two  adjacent  angles 
equal  to  two  right  angles,  these  two  form  one  straight 
line  GC  (corol.  1,  th.  6).  And  because  the  angle 
FBA  is  equal  to  the  angle  DBC,  being  each  a  right 
angle,  or  the  angle  of  a  square;  to  each  of  these 
equals  add  the  common  angle  ABC,  so  will  the  whole 
angle  or  sum  FBC  be  equal  to  the  whole  angle  or 
sum  ABD.  But  the  line  FB  is  equal  to  the  line  BA, 
being  sides  of  the  same  square  ;  and  the  line  BD  to 
the  line  BC,  for  the  same  reason;  so  that  the  two 
sides  FB,  BC,  and  the  included  angle  FBC,  are  equal 
to  the  two  sides  AB,  BD,  and  the  included  angle 
ABD,  each  to  each  ;  therefore  the  triangle  FBC  is 
equal  to  the  triangle  ABD  (th.  1). 

But  the  square  BG  is  double  the  triangle  FBC  on 


32  GEOMETRY. 

the  same  base  FB,  and  between  the  same  parallels 
FB,  GC  (th.  23) ;  in  like  manner,  the  parallelogram 
BL  is  double  the  triangle  ABD,  on  the  same  base 
BD,  and  between  the  same  parallels  BD,  AL.  And 
since  the  doubles  of  equal  things  are  equal  (by  ax.  0), 
therefore  the  square  BG  is  equal  to  the  parallelo- 
gram BL. 

In  like  manner,  the  other  square  CH  is  proved 
equal  to  the  other  parallelogram  EK.  Consequently, 
the  two  squares  BG  and  CH  together  are  equal  to 
the  two  parallelograms  BL  and  EK  together,  or  to 
the  whole  square  BE ;  that  is,  the  sum  of  the  two 
squares  on  the  two  less  sides  is  equal  to  the  square 
on  the  greatest  side.     Q.  E.  D. 

CoroL  1.  Hence  the  square  of  either  of  the  two 
less  sides  is  equal  to  the  difference  of  the  squares  of 
the  hypothenuse  and  the  other  side  (ax.  3) ;  or  equal 
to  the  rectangle  contained  by  the  sum  and  difference 
of  the  said  hvpothenuse  and  other  side;  for  (Alg.  13) 
«2-&2  =  (a  +  6)  (a-b). 

CoroL  2.  Hence,  also,  if  two  right-angled  triangles 
have  two  sides  of  the  one  equal  to  two  corresponding 
sides  of  the  other,  their  third  sides  will  also  be  equal, 
and  the  triangles  identical. 

THEOREM    XXVII.* 

In  any  triangle,  the  difference  of  the  squares  of  the 
two  sides  is  equal  to  the  difference  of  the  squares  of  the 
segments  of  the  base,  or  of  the  two  lines,  or  distances, 
included  between  the  extremes  of  the  base  and  the  per- 
pendicular. 

Let  ABC  be  any  triangle  hav-  c 

ing    CD    perpendicular    to   AB ; 
then   will   the   difference   of    the 
squares  of  AC,  BC  be  equal  to  the 
difference  of  the  squares  of  AD,  a       B   da 
BD ;  that  is,  AC2-BC2 = ADa-BD2. 

*  The  two  following  theorems  require  the  aid  of  the  following  al- 
gebraic formulas : 

(a-f  i)2  =  a2  +  2ai-f  b2  =a*  +  b*  -f  2ab 
(,a—by  =  a2—2abJrb2z=ai-\-b2—2ab 


THEOREMS.  33 

For,  since  ACD  and  BCD  are  right-angled  trian- 
gles, AC2=AD2+CD2)    ,      , 

and  BC2  =  BD2  +  CD2  \  W  tn-  26)  i 

.-.  By  subtraction,  AC2- BC2  =  AD2  — BD\ 

Corol.  The  rectangle  of  the  sum  and  difference  of 
the  two  sides  of  any  triangle  is  equal  to  the  rectangle 
of  the  sum  and  difference  of  the  distances  between 
the  perpendicular  and  the  two  extremes  of  the  base, 
or  equal  to  the  rectangle  of  the  base  and  the  differ- 
ence or  sum  of  the  segments,  according  as  the  per- 
pendicular falls  within  or  without  the  triangle. 

That  is,  (AC  +  BC) .  (AC  -  BC)  =  (AD  +  BD) . 
(AD-BD). 

Or,  (AC  +  BC) .  (AC  —  BC)  =  AB  (AD  -  BD)  in 
the  2d  fig. 

And,  (AC  +  BC) .  (AC  -  BC)  =  AB  .  (AD  -f-  BD) 
in  the  1st  fig. 

THEOREM    XXVIII. 

In  any  obtuse-angled  triangle,  the  square  of  the  side 
subtending  the  obtuse  angle  is  greater  than  the  sum  of 
the  squares  of  the  other  two  sides,  by  twice  the  rectangle 
of  one  of  the  sides  containing  the  obtuse  angle  and  the 
distance  of  the  perpendicular  drawn  from  the  opposite 
vertex  upon  this  side,  from  the  obtuse  angle. 

Let  ABC  be  a  triangle,  obtuse-angled  at  B,  and 
CD  perpendicular  to  AB ;  then  will  the  square  of 
AC  be  greater  than  the  squares  of  AB,  BC,  by  twice 
the  rectangle  of  AB,  BD.  That  is,  AC2  =  AB2  +  BC2 
+  2AB.  BD.     See  the  1st  fig.  above. 

For,  AD2  =  (AB  +  BD)2  =  AB2  +  BD2  +  2AB  . 
BD  ;  adding  CD2  to  both  members  of  this  equality, 
AD2  +  CD2  =  AB2  +  BD2  +  CD2  +  2AB  .  BD  (ax. 
2.) 

But  AD2  +  CD2  =  AC2,  and  BD2  +  CD2  =  BC2 
(th.  26). 

Therefore,  by  substitution  in  the  last  equality  but 
one,  AC2  -  AB2  -f  BC2  +  2AB  .  BD.     Q.  E.  D. 
B2 


34  oeumetrv. 


THEOREM    XXIX. 


In  any  triangle,  the  square  of  the  side  subtending 
an  acute  angle  is  less  than  the  squares  of  the  other  two 
sides,  by  twice  the  rectangle  of  one  of  the  sides  contain- 
ing the  acute  angle  and  the  distance  of  the  perpendicu- 
lar upon  this  side  from  the  acute  angle. 

Let  ABC  be  a  triangle,  having  c  c 

the  angle  A  acute,  and  CD  per-  a  a 

pendieular  to  AB  ;   then  will  the         /l\       /  j\ 
square  of  BC  be  less  than  the  sum      /      \     /    \\ 
of  the  squares  of  AB,  AC  by  twice  £- — -f-*  f — £-\ 
the  rectangle  of  AB,  AD  ;   that  is,  B   DA       D   B 

BC2  -  AB2  +  AC2  -  2  AD  .  AB. 

For  BD2=(AD  ~  AB)2  =  AD2  +  AB2  -  2AD  .  AB. 

And  BD2  +  DC2  -  AD2  +  DC2  -f  AB2  -  2AD .  AB 
(ax.  2). 

Therefore  BC2- AC2  +  AB2  -  2AD  .  AB  (th.  26). 
Q.  E.  D. 

THEOREM    XXX. 

In  any  triangle  the  double  of  the  square  of  a  line 
drawn  from  the  vertex  to  the  middle  of  the  base,  together 
with  double  the  square  of  the  half  base,,  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides. 

Let  ABC  be  a  triangle,  and  CD  the  c 

line  drawn  from  the  vertex  to  the  middle  , .-A 

of  the  base  AB,  bisecting  it  into  the  two  //  jV 

equal  parts  AD,  DB  ;  then  will  the  sum       /   I  \  \ 
of  the  squares  of  AC,  CB  be  equal  to  /       /    I    \ 
twice  the  sum  of  the  squares  of  CD,  AD  ;  A        DEB 
or  AC2  -f  CB2  =  2CD2  +  2AD2. 

For  AC2  =  CD2+  AD2  +  2AD  .  DE  (th.  28). 

And  BC2=  CD2  +  BD2  -  2 AD  .  DE  (th.  29). 

Therefore,  by  addition  (ax.  2), 
AC2  +  BC2  =  2CD2  +  AD2  +  BD2 

=  2CD2  +  2AD2  (by  hyp.).    Q.  E.  D. 


THEOREMS.  35 


THEOREM  XXXI. 

///  any  parallelogram  the  two  diagonals  bisect  each 
other,  and  the  sum  of  their  squares  is  equal  to  the  sum 
of  the  squares  of  all  the  four  sides  of  the  parallelogram. 

Let  ABDC  be  a  parallelogram 
whose  diagonals  intersect  each  other 
in  E ;  then  will  AE  be  equal  to  ED 
and  BE  to  EC,  and  the  sum  of  the 
squares  of  AD,  BC  will  be  equal  to       u  D 

the  sum  of  the  squares  of  AB,  BD,  CD,  CA  ;  that  is, 

AE  =  ED,  and  BE -EC, 
and      AD2  +  BC  =  AB2  +  BD2  +  CD2  +  CAa, 

For,  in  the  triangles  AEB,  DEC,  the  two  lines  AD, 
BC  meeting  the  parallels  AB,  DC,  make  the  angle 
BAE  equal  to  the  angle  CDE,  and  the  angle  ABE 
equal  to  the  angle  DCE,  and  the  side  AB  is  equal  to 
the  side  DC  (th.  19)  ;  therefore  these  two  triangles 
are  identical,  and  have  their  corresponding  sides 
equal  (th.  2),  viz.,  AE  =  DE,  and  BE  =  EC. 

Again,  since  AD  is  bisected  in  E,  the  sum  of  the 
squares,  CA3  +  CD2  =  2CEa  +  2DEa  (th.  30). 

In  like  manner,  BAa+  BD2=  2DE2-f  2BE2  or  2CE2. 

Therefore,  by  addition,  AB2  -f-  BD2  -f  DC2  +  CAa= 
4CE2  +  4DE2  (ax.  2). 

But  because  the  square  of  a  whole  line  is  equal  to 
4  times  the  square  of  half  the  line  ;*  that  is,  BCa= 
4BE2,  and  AD'  =  4DEa; 

Therefore  AB2  -f  BD2  -f-  DC2  -f  C  A2  =  BC2  -f  ADa 
(ax.  1).     Q.  E.  D. 

Cor.  If  AB  =  AC,  or  the  parallelogram  be  a  rhom- 
bus, then  the  triangles  AEB,  AEC  will  be  mutually 
equilateral,  and,  consequently  (th.  5),  the  angle  BEA 
of  the  one  will  be  equal  to  the  angle  AEC  of  the 
other.  Hence  (def.  12)  the  diagonals  of  a  rhombus 
intersect  at  right  angles. 


This  may  be  seen  from  the  accompanying  diagram,  or, 
algebraically,  from  considering  that  the  square  of  la  is  \a*. 


36  ueomktkv. 


EXERCISES. 

1.  To  construct  an  isosceles  triangle  with  a  given  base  and  given 
vertical  angle. 

2.  Prove  that  every  point  of  the  bisectrix  of  a  given  angle  is  equally 
distant  from  the  sides. 

3.  Two  angles  of  a  triangle  being  given,  to  find  the  third. 

4.  To  construct  an  isosceles  triangle  so  that  the  vertex  shall  fall  at 
a  given  point,  and  the  base  fall  in  a  given  line. 

5.  An  isosceles  triangle  so  that  the  base  shall  be  a  given  line  and 
the  vertical  angle  a  right  angle. 

6.  With  two  angles  and  a  side  opposite  one  of  them,  to  construct 
a  triangle. 

7.  To  construct  a  triangle  when  the  base,  the  angle  opposite,  and 
the  sum  of  this  and  one  of  the  other  two  angles  are  given. 

8.  The  same,  except  the  difference  instead  of  the  sum  given. 

9.  To  construct  a  quadrilateral  when  the  four  sides  and  one  angle 
are  given. 

10.  When  three  of  the  sides  and  the  two  angles  included  between 
them  are  given. 

11.  When  two  sides  and  the  included  angle  and  two  other  angles. 

12.  To  construct  a  parallelogram  with  two  adjacent  6ides  and  the 
diagonal  given. 

13.  To  construct  a  parallelogram  with  given  base,  altitude,  and 
diagonal. 

14.  With  two  adjacent  sides  and  the  altitude. 

15.  To  make  a  hexagon  equal  in  all  respects  to  a  given  irregular 
hexagon. 

16.  To  construct  a  triangle  with  the  angles  at  the  base  and  the  alti- 
tude given. 

17.  With  the  vertical  angle,  one  of  its  sides  and  the  altitude  given. 

18.  With  the  base,  altitude,  and  one  of  the  angles  at  the  base  given. 

19.  To  construct  a  trapezoid  when  three  sides  and  the  angle  con- 
tained between  two  of  them  are  given. 

20.  A  line  and  two  points  without  it  being  given,  to  find  a  point  in 
the  line  equidistant  from  the  two  given  points.* 

*  Geometi-ic  Analysis. — The  best  method  for  discovering  the  solu- 
tion of  problems  is  what  is  termed  the  analytic.  This  consists  in 
supposing  the  problem  solved,  making  the  diagram  accordingly,  and. 
then,  by  examination  of  the  required  and  given  parts  of  the  diagram 
in  their  relations  to  one  another,  considering  what  known  theorems 
of  geometry  connect  them  together.  This  is  a  sort  of  going  back 
from  the  result  sought  by  a  chain  of  relations — depending  upon  known 


EXERCISES.  37 

21.  TUe  data  as  above  to  draw  two  lines  from  the  two  given  points, 
meeting  in  the  given  line,  and  making  equal  angles  with  it. 

theorems  to  what  is  given  (or  may  be  obtained),  and  is  the  natural 
m  of  discovery  or  invention. 

The  required  result  having  been  thus  obtained  by  analysis,  or  reso- 
lution, the  demonstration  of  its  correctness  is  made  by  synthesis,  or 
composition;  the  order  in  which  is  the  reverse  of  the  former,  and 
Carries  us  forward  from  the  data,  by  means  of  the  truths  on  which  the 
rmtilt  depend*,  to  the  result  itself.  Analysis  is,  then,  the  method  of 
discovery,  synthesis  of  demonstration  after  the  discovery  is  made.  The 
one  has  for  its  object  to  find  unknown  truths,  the  other  to  prove 
known  ones.  Analysis  and  synthesis  are  both  of  them  applicable  to 
theorems  as  well  as  problems.  In  submitting  a  problem  to  analysis,  its 
solution,  in  the  first  instance,  is  assumed ;  and  from  this  assumption  a 
series  of  consequences  are  drawn,  until  at  length  something  is  found 
which  can  be  done  upon  established  principles.  In  the  synthesis,  or 
solution,  beginning  with  the  construction  indicated  by  the  final  result 
of  the  analysis,  the  process  ends  with  the  performance  of  what  was 
required  by  the  problem,  and  is  the  first  step  of  the  analysis. 

When  a  theorem  is  submitted  to  analysis,  the  thing  to  be  determ- 
ined is  whether  the  statement  expressed  by  it  be  true  or  not.  In 
the  analysis  this  statement  is,  in  the  first  instance,  assumed  to  be  true, 
and  a  series  of  consequences  deduced  from  it,  until  some  result  is  ob- 
tained, which  either  is  an  established  or  admitted  truth,  or  contradicts 
an  established  or  admitted  truth.  If  the  former,  the  theorem  may  be 
proved  by  retracing  the  steps  of  the  investigation,  commencing  with 
the  final  result  and  concluding  with  the  proposed  theorem.  But  if 
the  final  result  contradict  an  established  truth,  the  proposed  theorem 
must  be  false,  since  it  leads  to  a  false  conclusion. 

We  give  a  specimen  of  the  analytic  investigation  of  a  problem  be- 
low. Of  synthesis  the  student  has  already  had  specimens  in  all  the 
preceding  theorems,  and  will  find  others  in  the  problems  which  fol- 
low the  remaining  theorems  of  plane  geometry. 

Specimen  of  the  Analysis  of  a  Problem. 

Given  two  angles,  and  the  sum 
of  the  three  sides  of  a  triangle,  to 
construct  it. 

Suppose  it  done,  and  that  ABO 
is  the  triangle  sought.  Produce 
BC  till  CD  =  CA  and  BE  =  BA; 
join  EA,  DA;  then  the  triangles 
ACD  and  ABE  being  isosceles,  the  angle  ABC  =  twice  the  angle 
AEB,  and  the  angle  ACB  =  twice  the  angle  ADC. 

Hence  the  following  construction:  at  the  extremities  of  a  line  ED, 
equal  to  the  given  sum  of  the  three  sides,  draw  lines  making,  with 
this,  angles  each  equal  to  half  one  of  the  given  angles,  and  from 
the  point  A,  where  they  meet,  draw  lines  making  angles  with  AE 
gad  A  I),  respectively  equal  to  the  angles  E  and  D ;  ABC  will  be  the 
triangle  required. 

The  demonstration  synthetically  would  be  as  follows: 

The  angle  EAB  being  =  the  angle  E,  the  triangle  is  isosceles,  and 


38  GEOMETRY. 

22.  The  same  when  the  two  given  points  are  on  opposite  sides  of 
the  line. 

23.  When  every  side  of  a  polygon  is  produced  out,  prove  that  the 
sum  of  the  outward  angles  is  equal  to  four  right  angles. 

24.  Show  that  in  an  isosceles  triangle  the  square  of  the  line  drawn 
from  the  vertex  to  any  point  of  the  base,  together  with  the  rectangle 
of  the  segments  of  the  base,  is  equal  to  the  square  of  one  of  the  equal 
sides  of  the  triangle. 

25.  Prove  that  the  square  of  a  line  is  equal  to  the  square  of  its  pro- 
jection on  another  line  added  to  the  square  of  the  difference  of  the 
perpendiculars  which  determine  this  projection. 

26.  Prove  that  the  sum  of  the  squares  on  two  lines,  together  with 
twice  their  rectangle,  is  equal  to  the  square  on  their  sum. 

27.  That  the  square  on  the  difference  of  two  lines  is  equal  to  the 
sum  of  their  squares  minus  twice  their  rectangle. 

28.  To  construct  a  quadrangle  when  three  sides,  one  angle,  and 
the  sum  of  two  other  angles  are  given. 

29.  When  three  angles  and  two  opposite  sides. 

30.  Prove  that  two  parallelograms  are  equal  when  they  have  two 
sides  and.  the  included  angle  equal. 

31.  Prove  that  the  greater  diagonal  of  a  parallelogram  is  opposite 
the  greater  angle. 

32.  Prove  that  two  rhombi  are  equal  when  a  side  and  angle  of 
the  one  are  equal  to  the  same  in  the  other. 

33.  That  if  the  diagonals  of  a  quadrilateral  bisect  each  other  at 
right  angles,  the  figure  will  be  a  rhombus. 

34.  That  the  diagonals  of  a  rectangle  are  equal;  and  the  converse. 

35.  Prove  that  the  line  joining  the  middle  points  of  the  inclined 
sides  of  a  trapezoid  is  parallel  to  the  bases,  and  that  it  is  equal  to 
half  the  sum  of  the  bases. 

36.  Prove  that  two  convex  polygons  are  identical:  1°.  When  they 
have  the  same  vertices.  2°.  When  one  side  in  each  equal,  and  the 
distances  of  the  corresponding  vertices  from  its  extremities  equal. 
3°.  When  composed  of  the  same  number  of  equal  triangles,  similarly 
placed.  4°.  When  they  have  all  their  sides  equal  and  all  their  an- 
gles but  two.  5°.  When  all  their  sides  but  one  and  all  their  angles 
but  one. 

37.  Prove  that  there  can  be  but  one  perpendicular  from  a  given 
point  to  a  given  line. 

AB  =  BE;  for  a  similar  reason  AC  =  CD;  hence  the  sum  of  the 
three  sides  of  the  triangle  ABC  =  the  given  sum  ED.  Again,  the 
angle  ABC  =  2AEB,  and  ACB  =  2ADC  .-.  ABC  and  ACB  are  equal 
the  given  angles.     Q.  E.  D. 


THEOREMS.  39 


THEOREM  XXXII. 


If  two  triangles  have  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  but  the  included  angles  unequal, 
the  third  sides  will  be  unequal,  and  the  greater  will 
be  in  that  triangle  which  has  the  greater  included 
angle. 

Let  ABC,  DEF 
be  two  triangles  in 
which  AB  =  DE, 
AC  =  DF,  BAC  < 
EDF.      Then   will 


EF  >  BC.  B  G  ~         ^-Ag 

For  at  the  point  D  make  the  angle  EDG  equal  to 
the  angle  BAC  ;  take  DG  equal  to  AC,  and  join  GE. 
Then  will  the  triangle  DEG  equal  the  triangle  ABC 
(th.  1)  and  EG  =  BC. 

But  EG  <  EI  +  IG,  and  DF  <  DI  -f  if  (ax.  13). 

By  addition  of  these  inequalities,  EG  4-  DF  <  EI 
+  IF  +  DI  +  IG ;  or,  EG  +  DF  <  EF  +  DG. 

Taking  away  the  equals  DF  and  DG  from  the 
members  of  the  last  inequality,  there  remains  EG  < 
EF.     But  EG  =  BC  .-.  BC  <  EF.     Q.  E.  D. 

If  the  point  G  fall  within  instead  of  without  the 
triangle  DEF,  we  should  have  DG  +  GE  <  DF  + 
EF  (th.  14)  ;  and,  taking  away  the  equals  DG  and 
DF,  there  remains  EG  <  EF.  If  the  point  G  fall  on 
EF,  the  theorem  is  evident. 

The  converse  of  this  proposition  is  also  true,  viz., 
that  if  two  sides  of  one  triangle  be  equal  to  two  sides 
of  another,  and  the  third  sides  unequal,  the  angle  op- 
posite the  smaller  third  side  will  be  less  than  the 
one  opposite  the  larger.  For  if  the  angle  were  great- 
er, by  the  above  proposition  the  third  side  must  be 
greater;  and  if  it  were  equal,  it  must,  by  (th.  1),  be 
equal.  But  the  third  side  is  neither  greater  nor  equal ; 
therefore  the  angle  opposite,  being  neither  greater 
nor  equal  than  the  angle  of  the  other  triangle,  must 
be  less. 


40  GEOMETRY. 


THEOREM    XXXIII. 

Every  diameter  bisects  a  circle  and  its  circumference. 

Let  ACBD  be  a  circle,  AB  a  di- 
ameter. Conceive  the  part  ADB 
to  be  turned  over  and  applied  to  the 
part  ACB,  it  will  coincide  with  it  .  / 
exactly,  otherwise  there  would  be 
points  in  the  one  portion  of  the  cir- 
cumference or  the  other  unequally 
distant  from  the  center;  but  this  is  D 

contrary  to  the  definition  (def.  41)  ;   hence  the  two 
parts  are  equal  (ax.  10). 

THEOREM    XXXIV. 

If  a  line  drawn  through  the  center  of  a  circle  bisect 
a  chord,  it  tvill  be  pe?~pendicular  Ho  the  chord ;  or,  if 
it  be  perpendicular  to  the  chord,  it  will  bisect  both  the 
chord  and  the  arc  of  the  chord. 

Let  AB  be  any  chord  in  a  circle, 
and  CD  a  line  drawn  from  the  cen- 
ter C  to  the  chord.  Then,  if  the 
chord  be  bisected  in  the  point  D, 
CD  will  be  perpendicular  to  AB. 

Draw  the  two  radii  CA,  CB. 
Then  the  two  triangles  ACD,  BCD, 
having  CA  equal  to  CB,  being  ra- 
dii of  the  same  circle  (def.  41),  and  CD  common,  also 
AD  equal  to  DB  (by  hyp.)  ;  they  have  all  the  three 
sides  of  the  one  equal  to  all  the  three  sides  of  the 
other,  and  so  have  their  angles  also  equal  (th.  5). 
Hence,  then,  the  angle  ADC  being  equal  to  the  angle 
BDC,  these  angles  are  right  angles,  and  the  line  CD 
is  perpendicular  to  AB  (def.  12). 

Again,  if  CD  be  perpendicular  to  AB,  then  will  the 
chord  AB  be  bisected  at  the  point  D,  or  have  AD 
equal  to  DB ;  and  the  arc  AEB  bisected  in  the  point 
E,  or  have  AE  equal  EB. 


TMBOfcCMft.  41 

For,  the  two  triangles  ACD,  BCD  being  right-an- 
bled  at  D,  and  having  two  sides  of  the  one  equal  tG 
the  same  in  the  other,  viz.,  AC  =  CB  and  CD  common, 
are  equal  (th.  26,  cor.  2),  .-.  AD^=BD. 

Also,  since  the  angle  ACE  is  equal  to  the  angle 
BCE,  the  arc  AE,  which  measures  the  former,  is  equal 
to  the  arc  BE,  which  measures  the  latter,  since  equal 
angles  must  have  equal  measures. 

Scholium.  Two  conditions  determine  a  line  such  as 
that  it  shall  pass  through  two  given  points,  or  that  it 
shall  pass  through  one  point  and  be  perpendicular  to 
a  given  line,  or  pass  through  a  point  and  make  a  given 
angle  with  a  given  line. 

The  line  CE  in  the  last  diagram  fulfills  four  condi- 
tions. It  passes  through  the  center  C,  through  the 
point  D,  the  middle  of  the  chord,  through  the  point 
E,  the  middle  of  the  arc,  and,  finally,  is  perpendicular 
to  the  chord  AB.  Either  two  of  these  involves  the 
other  two.  Thus,  if  a  line  pass  through  the  middle  of 
the  chord  and  be  perpendicular  to  it,  it  will  pass 
through  the  middle  of  the  arc  and  the  center  of  the 
circle ;  if  it  pass  through,  the  middle  of  the  arc  and 
center  of  the  circle,  it  will  pass  through  the  middle 
of  the  chord  and  be  perpendicular  to  it ;  if  it  pass 
through  the  middle  of  the  arc  and  chord,  it  will  be 
perpendicular  to  the  latter,  and  pass  through  the  cen- 
ter of  the  circle,  &c. 

•*% 

THEOREM    XXXV. 

Any  chords  in  a  circle  which  are  equally  distant 
from  the  center  are  equal  to  each  other  ;  or,  if  they  be 
equal  to  each  other,  they  will  be  equally  distant  from 
the  center. 

Let  AB,  CD  be  any  two  chords  at 
equal  distances  from  the  center  G  ;  then 
will  these  two  chords  AB,  CD  be  equal 
to  each  other. 

Draw  the  two  radii  GA,  GC,  and  the 
two  perpendiculars  GE,  GF,  which  are 


42  GEOMETRY. 

the  equal  distances  of  AB,  CD  from  G  (th.  17).*  Then 
the  two  right-angled  triangles,  GAE,  GCF,  having  the 
side  GA  equal  the  side  GC  (being  radii),  and  the  side 
GE  equal  the  side  GF,  are  identical  (cor.  2,  th.  26),  and 
have  the  line  AE  equal  to  the  line  CF.  But  AB  is 
the  double  of  AE  (th.  34),  and  CD  is  the  double  of 
CF  ;  therefore  AB  is  equal  to  CD  (by  ax.  6).     Q.  E.  D. 

Again,  if  the  chord  AB  be  equal  to  the  chord  CD, 
then  will  their  distances  from  the  center,  GE,  GF, 
also  be  equal  to  each  other. 

For,  since  in  the  right-angled  triangles  AEG,  CFG, 
AE  the  half  of  AB  is  equal  to  CF,  the  half  of  CD,  and 
the  radii  GA,  GC  are  equal,  therefore  the  third  sides 
are  equal  (cor.  2,  th.  26),  or  the  distance  GE  is  equal 
the  distance  GF.     Q.  E.  D. 

THEOREM    XXXVI. 

A  line  perpendicular  to  a  radius  at  its  extremity  is 
a  tangent  to  the  circle. 

Let  the  line  ADB  be  perpendicular  A  ^  ^  E  B 
to  the  radius  CD  of  a  circle  ;  then  is 
any  other  point,  except  D,  as  E  of  the 
line  AB,  without  the  circle.  For  CE, 
an  oblique  line,  is  greater  than  the  per- 
pendicular CD  (th.  17),  or  greater  than  the  radius. 
Hence,  the  line  AB  having  but  one  point,  D,  in  com- 
mon with  the  circle,  is  a  tangent  (def.  56). 

THEOREM    XXXVII. 

When  a  line  is  a  tangent  to  a  circle,  a  radius  drawn 
to  the  point  of  contact  is  perpendicular  to  the  tangent. 

For  if  oblique,  a  line  shorter  can  be  drawn  perpen- 
dicular to  the  tangent,  and  the  tangent  must  then 
pass  within  the  circle,  which  is  contrary  to  definition. 

Corol.  1.  Hence,  conversely,  a  line  drawn  perpen- 
dicular to  a  tangent,  at  the  point  of  contact,  passes 
through  the  center  of  the  circle  ;  for  there  can  be  but 
one  perpendicular  to  a  given  line  through  a  given  point. 

*  By  the  distance  of  a  poiut  from  a  line  is  understood  the  shortest 
distance. 


THEOREMS.  43 

Corol.  2.  If  any  number  of  circles  touch 
each  other  at  the  same  point,  their  centers 
must  be  in  the  same  line  perpendicular  to 
their  common  tangent  ;  for  the  perpen- 
dicular to  the  tangent  at  the  common  point 
must  pass  through  the  center  of  each. 

THEOREM    XXXVIII. 

The  angle  formed  by  a  tangent  and  chord  is  meas- 
ured by  half  the  arc  of  that  chord. 

Let  AB  be  a  tangent  to  a  circle,  and  A_    C       B 
CD  a  chord  drawn  from  the  point  of 
contact  C  ;  then  is  the  angle  BCD  meas-    / 
ured  by  half  the  arc  CFD,  and  the  angle    [ 
ACD  measured  by  half  the  arc  CGD.      G< 

Draw  the  radius  EC  to  the  point  of 
contact,  and  the  radius  EF  perpendicular  to  the  chord 
atH. 

Then  the  radius  EF,  being  perpendicular  to  the 
chord  CD,  bisects  the  arc  CFD  (th.  34).  Therefore 
CF  is  half  the  arc  CFD. 

But  the  angle  CEF  is  equal  to  the  angle  BCD,  be- 
cause the  sides  of  the  two  angles  are  respectively 
perpendicular  to  each  other,  and  consequently  have 
the  same  difference  of  direction.  Moreover,  the  angle 
CEF  is  measured  by  the  arc  CF  (def.  10,  note),  which 
is  the  half  of  CFD ;  therefore  the  equal  angle  BCD 
must  also  have  the  same  measure,  namely,  half  the 
arc  CFD  of  the  chord  CD. 

Again,  GEF  being  a  diameter,  CG  is  the  supple- 
ment of  CF,  and  is  equal  to  GD,  the  supplement  of  FD. 
.*.  CG  is  half  the  arc  CGD.  Now,  since  the  line  CE, 
meeting  FG,  makes  the  sum  of  the  two  angles  at  E 
equal  to  two  right  angles  (th.  6),  and  the  line  CD 
makes  with  AB  the  sum  of  the  two  angles  at  C  equal 
to  two  right  angles  ;  if  from  these  two  equal  sums 
there  be  taken  away  the  parts  or  angles  CEH  and 
BC1I,  which  have  been  proved  equal,  there  remains 
the  angle  CEG,  equal  to  the  angle  ACH.     But  the 


44  GEOMETRY. 

former  of  these,  CEG,  being  an  angle  at  the  center, 
is  measured  by  the  arc  CG  (def.  10,  note) ;  conse- 
quently the  equal  angle  ACD  must  also  have  the  same 
measure  CG,  which  is  half  the  arc  CGD  of  the  chord 
CD.     Q.  E.  D. 

.  CoroL  1.  The  sum  of  two  right  angles  is  measured 
by  half  the  circumference.  For  the  two  angles  BCD, 
ACD,  which  make  up  two  right  angles,  are  measured 
by  the  arcs  CF,  CG,  which  make  up  half  the  circum- 
ference, FG  being  a  diameter. 

Coral.  2.  Hence,  also,  one  right  angle  must  have 
for  its  measure  a  quarter  of  the  circumference,  or  90 
degrees. 

THEOREM  XXXIX. 

An  angle  at  the  circumference  of  a  circle  is  measur- 
ed by  half  the  arc  that  subtends  it. 

Let  B  AC  be  an  angle  of  the  circum-  D 
ference  ;  it  has  for  its  measure  half  the 
arc  which  subtends  it. 

For,  suppose  the  tangent  DE  to  B\ 
pass  through  the  point  of  contact  A ; 
then,  the  angle  DAC  being  measured  by  half  the  arc 
ABC,  and  the  angle  DAB  by  half  the  arc  AB  (th.  38), 
it  follows,  by  equal  subtraction,  that  the  difference,  or 
angle  BAC,  must  be  measured  by  half  the  arc  BC, 
upon  which  it  stands.     Q.  E.  D. 

CoroL  1.  All  angles  in  the  same  segment  of  a  circle, 
or  standing  on  the  same  arc,  are  equal  to  each  other. 

CoroL  2.  An  angle  at  the  center  of  a  circle  is  double 
the  angle  at  the  circumference,  when  both  stand  on 
the  same  arc. 

CoroL  3.  An  angle  in  a  semicircle  is  a  right  angle. 

THEOREM    XL. 

Any  two  parallel  chords  intercept  equal  arcs. 

Let  the  two  chords  AB,  CD  be  parallel ;  then  will 
the  arcs  AC,  BD  be  equal ;  or  AC  =  BD. 


THEOREM*.  45 

Draw  the  line  BC.  Then,  because  theA 
lines  AB,  CD  are  parallel,  the  alternate  an- 
gles B  and  C  are  equal  (th.  10).  But  the 
angle  at  the  circumference  B  is  measured0 
by  half  the  arc  AC  (th.  39)  ;  and  the  other  equal  an- 
gle at  the  circumference  C  is  measured  by  half  the  arc 
BD  ;  therefore  the  halves  of  the  arcs  AC,  BD,  and 
consequently  the  arcs  themselves,  are  also  equal.  Q. 
E.D. 


THEOREM    XLI. 

When  a  tangent  and  chord  are  parallel  to  each  other, 
they  intercept  equal  arcs. 

Let  the  tangent  ABC  be  parallel  to 
the  chord  DF;  then  are  the  arcs  BD, 
BF  equal ;  that  is,  BD  =  BF. 

Draw  the  chord  BD.  Then,  because 
the  lines  AB,  DF  are  parallel,  the  al- 
ternate angles  D  and  B  are  equal  (th.  10).  But  the 
angle  B,  formed  by  a  tangent  and  chord,  is  measured 
by  half  the  arc  BD  (th.  38) ;  and  the  other  angle  at 
the  circumference  D  is  measured  by  half  the  arc  BF 
(tli.  39)  ;  therefore  the  arcs  BD,  BF  are  equal.  Q. 
E.  D. 

THEOREM    XLII. 

When  two  lines,  meeting  a  circle  each  in  two  points, 
cut  one  another,  either  within  or  without  it,  the  rectangle 
of  the  parts  of  the  one  is  equal  to  the  rectangle  of  the 
marts  of  the  other*  the  parts  of  each  being  measured 
from  the  point  of  meeting  to  the  two  intersections  with 
the  circumference. 

Let  the  two  lines  AB,  CD  meet  each 
other  in  E  ;  then  the  rectangle  of  AE,  EB 
will  he  equal  to  the  rectangle  of  CE,  ED. 
Or,  AE  .  EB  =  CE  .  ED. 

For  through  the  poiut  E  draw  thevliame- 


46  GEOMETRY. 

tcr  FG;  also,  from  the  center  H  draw  the         a 
radius  DH,  and  draw  HI  perpendicular  to      c 

CD.  /ffi\ 
Then,  since  DEH  is  a  triangle,  and  the  (A>iH\ ) 

perpendicular  HI  bisects  the  chord  CD  (th.  D\^  \^Sb 
34),  the  line  CE  is  equal  to  the  difference  of  G 
the  segments  DI,  EI,  the  sum  of  them  being  DE.  Also, 
because  H  is  the  center  of  the  circle,  and  the  radii  DH, 
FH,  GH  are  all  equal,  the  line  EG  is  equal  to  the 
sum  of  the  sides  DH,  HE ;  and  EF  is  equal  to  their 
difference. 

But  the  rectangle  of  the  sum  and  difference  of  the 
two  sides  of  a  triangle  is  equal  to  the  rectangle  of  the 
sum  and  difference  of  the  segments  of  the  base  (cor., 
th.  27) ;  therefore  the  rectangle  of  FE,  EG  is  equal  to 
the  rectangle  of  CE,  ED.  In  like  manner,  it  is  proved 
that  the  same  rectangle  of  FE,  EG  is  equal  to  the 
rectangle  of  AE,  EB.  Consequently,  the  rectangle 
of  AE,  EB  is  also  equal  to  the  rectangle  of  CE,  ED 
(ax.  1).     Q.  E.  D.  v 

Corol.  1.  When  one  of  the  lines  in  the 
second  case,  as  DE,  by  revolving  about    c 
the  point  E,  comes  into  the  position  of  the      ' 
tangent  EC  or  ED,  the  two  points  C  and 
D  running  into  one  ;  then  the  rectangle  of 

CE,  ED  becomes  the  square  of  CE,  be- 
cause CE  and  DE  are  then  equal.  Consequently,  the 
rectangle  of  the  parts  of  the  secant,  AE  .  EB,  is  equal 
to  the  square  of  the  tangent,  CE\ 

Corol.  2.  Hence  both  the  tangents  EC,  EF,  drawn 
from  the  same  point  E,  are  equal ;  since  the  square 
of  each  is  equal  to  the  same  rectangle  or  quantity 
AE .  EB. 

THEOREM    XLIII. 

In  equiangular  triangles,  the  rectangles  of  the  cor- 
responding or  like  sides,  taken  alternately,  are  equal. 

Let  ABC,  DEF  be  two  equiangular  triangles, 
having  the  angle  A  =  the  angle  D,  the  angle  B  *=■ 


THEOREMS.  47 

the  angle  E,  and  the  angle  C  =  the 
angle  F ;  also,  the  like  sides  AB,  DE, 
and  AC,  DF,  being  those  opposite  the 
equal  angles;  then  will  the  rectangle 
of  AB,  DF  be  equal  to  the  rectangle 
of  AC,  DE. 

In  BA,  produced,  take  AG  equal  to  DF ;  and  through 
the  three  points  B,  C,  G,  conceive  a  circle  BCGH  to  be 
described,  meeting  CA,  produced,  at  H,  and  join  GH. 

Then  the  angle  G  is  equal  to  the  angle  C  on  the 
same  arc  BH,  and  the  angle  H  equal  to  the  angle  B 
on  the  same  arc  CG  (th.  39) ;  also,  the  opposite  angles 
at  A  are  equal  (th.  7) :  therefore  the  triangle  AGH  is 
equiangular  to  the  triangle  ACB,  and  consequently  to 
the  triangle  DFE  also.  But  the  two  like  sides  AG, 
DF  are  also  equal,  by  construction;  consequently,  the 
two  triangles  AGH,  DFE  are  identical  (th.  2)",  and 
have  the  two  sides  AG,  AH  equal  to  the  two  DF, 
DE,  each  to  each. 

But  the  rectangle  GA  .  AB  is  equal  to  the  rectangle 
HA  .  AC  (th.  42)  ;  consequently,  the  rectangle  DF  . 
AB  is  equal  to  the  rectangle  DE  .  AC.     Q.  E.  D. 

KXKUCISES. 

1.  Find  the  length  of  an  arc  of  20°  45'  to  a  radius  of  10. 

2.  Through  two  given  points,  to  draw  a  circumference  of  given 
radius. 

3.  Divide  an  arc  into  2,  4,  8,  16  ... .  equal  parts. 

4.  Prove  that  every  other  chord  is  less  than  the  diameter. 

5.  That  parallel  tangents  include  semicircumferences  between  their 
points  of  contact. 

6.  Draw  a  tangent  to  a  given  circle  parallel  to  a  given  line. 

7.  To  describe  a  circle  of  given  radius  tangent  to  a  given  line  at  a 
given  point 

8.  To  describe  a  circle  of  given  radius  touching  the  two  sides  of  a 
given  angle. 

9.  To  describe  a  circumference  which  shall  be  embraced  between 
two  parallels,  and  pass  through  a  given  point. 

10.  To  place  a  chord  of  given  length  and  direction  in  a  given  circle. 

1 1 .  Prove  that  the  chords  of  equal  arcs  are  equal,  and  the  converse. 

12.  To  find  in  one  side  of  a  triangle  the  center  of  a  circle  which 
shall  touch  the  other  two  sides. 


48  GEOMETRY. 

13.  To  find  the  radius  of  a  circle  when  a  chord  and  perpendicular 
from  the  centre  to  the  chord  are  given. 

14.  With  given  radii  to  describe  two  circumferences  which  shall 
intersect  in  a  given  point,  and  have  their  centers  in  a  given  line. 

15.  With  given  radii  to  describe  two  circles  which  shall  touch  each 
other  either  externally  or  internally. 

16.  Three  circles  with  equal  given  radii  touching  each  other  ex- 
ternally. 

17.  The  same  with  unequal  radii. 

18.  Through  a  given  point  on  a  circumference,  and  another  given 
point  without,  to  describe  a  circle  touching  the  given  circumference. 

19.  The  same  when,  instead  of  the  point  upon  the  circumference, 
the  radius  of  the  required  circle  is  given. 

20.  To  describe  a  circle  of  given  radius  touching  two  given  circles. 

21.  To  construct  a  right-angled  triangle  with  the  hypothenuse  and 
one  of  the  perpendicular  sides  given. 

22.  In  a  given  circle  to  inscribe  a  right  angle,  one  side  of  which  is 
given. 

23.  In  a  given  circle  to  construct  an  inscribed  triangle  of  given 
altitude  and  vertical  angle. 

24.  Also,  a  quadrangle,  when  one  side  and  two  angles  not  adjacent 
this  side  are  given.     (See  Exercise  31,  below.) 

25.  To  find  the  center  of  a  circle  in  which  two  given  lines  meet- 
ing in  a  point  shall  be  a  tangent  and  chord. 

26.  In  a  given  circle  to  inscribe  a  triangle  equiangular  to  a  given 
triangle. 

27.  Show  how  to  circumscribe  a  square  about  a  given  circle,  and 
how  to  inscribe  a  circle  in  a  given  square. 

28.  That  a  straight  line  touching  a  circle  can  have  with  it  but  one 
point  of  contact. 

29.  To  inscribe  in  an  equilateral  triangle  three  equal  circles  touch- 
ing each  other,  and  the  sides  of  the  triangle. 

30.  Prove  that  an  eccentric  angle  is  measured  by  half  the  sum  of 
the  opposite  arcs  subtending  it,  if  the  vertex  be  within  the  circle  ; 
and  by  half  the  difference  of  the  arcs  if  it  be  without. 

31.  That  the  opposite  angles  of  an  inscribed  quadrilateral  are  sup- 
plements. 

32.  That  if  one  of  the  sides  of  an  inscribed  quadrilateral  be  produced 
out,  the  outward  angle  will  be  equal  to  the  inward  opposite  angle. 

33.  That  the  sums  of  the  opposite  sides  are  equal. 

34.  That  a  regular  polygon  may  be  circumscribed  with  a  circle. 

35.  That  a  circle  may  be  inscribed  in  any  regular  polygon. 


EXERCISES.  49 

36.  If  one  circle  touch  another  externally  or  internally,  any  straight 
lint  drawn  through  the  point  of  contact  will  cut  off  similar  tegmenta." 

37.  Prove  that  only  one  tangent  can  he  drawn  to  a  circle  at  a 
given  point  on  the  circumference. 

38.  That  of  two  chords  the  greater  is  nearer  the  center  of  the  circle. 

Numerical  Problems. 
1.  In  a  triangle  suppose  two  of  the  sides  to  be  8.76  and  5.26,  and 
the  perpendicular  from  the  vertex  in  which  they  meet  4.38;  required 
the  third  side. 

Suppose  the  two  segments  of  the  required  side  to  be  represented 
by  x  and  y. 

x  —  ^/ (8.76)3  —  (4^38)*,  y  =  ^(5.26)*—  (4.38)*, 
or,  •* 


x  =  v/  (8.76  +  4.38)  (8.76  —  4.38),  y  =  y/  (5.26  -f  4.38)  (5.26  —  4.38). 

log.  13.14  =  1.1185954 

log.    4.38  =  0.6414741 

2)1.7600695 

x  —  7.586,  log.  =  0.8800347 

By  a  similar  method,  y  =  2.9126 

.-.x-fy=  10.4990 
for  the  value  of  the  third  side  if  the  perpendicular  falls  within  the 
triangle;  and  x^,y  =  4.67 

for  the  value  if  the  perpendicular  falls  without. 

itren  in  a  triangle  the  base  88;  one  of  the  sides  128.49;  and 
the  perpendicular  upon  the  base  from  the  vertex  opposite  96.45,  to 
iind  the  third  side.     Ans.  96.50. 

3.  Two  chords  cut  each  other  in  a  circle;  the  segments  of  the  one 
are  13  and  25 ;  the  segments  of  the  other  are  in  the  ratio  of  4  to  7  ; 
required  the  length  of  the  latter  chord.     Ans.  37.47. 

4.  To  find  the  absolute  length  of  an  arc  of  45°  20'  in  a  circle  whoso 
rail  ins  is  5.4,  supposing  the  ratio  of  the  circumference  to  the  diameter 
of  :i  circle  to  be  3.1416.     Ans.  4.2726. 

5.  The  side  of  a  square  being  given  0.25,  to  find  the  side  of  an 
equilateral  triangle  equal  to  the  square.     Ans.  0.37994. 

6.  Given  the  area  of  a  circle  33.1830,  to  find  its  radius.     Ans.  3.25. 

7.  Find  the  chord  of  the  sum  of  two  arcs,  the  chords  of  the  arcs 
being  given  10  and  12,  and  the  radius  16. 

8.  Find  the  chord  of  half  an  arc,  the  chord  of  the  whole  arc  being 
12  and  16. 

*  Similar  segments  are  those  which  correspond  to  similar  arcs. 

c 


OF  RATIOS  AND  PROPORTIONS. 


DEFINITIONS. 

Def.  75.  Ratio  is  the  proportion  or  relation  which 
one  magnitude  bears  to  another  magnitude  of  the 
same  kind  with  respect  to  quantity. 

Note. — The  measure  or  quantity  of  a  ratio  is  con- 
ceived by  considering  what  part  or  parts  the  leading 
quantity,  called  the  Antecedent,  is  of  the  other,  called 
the  Consequent  ;*  or  what  part  or  parts  the  number 
expressing  the  quantity  of  the  former  is  of  the  num- 
ber denoting,  in  like  manner,  the  latter.  So  the  ratio 
of  a  quantity,  expressed  by  the  number  2  to  a  like 
quantity  expressed  by  the  number  6,  is  denoted  by  2 
divided  by  6,  or  f  or  £ ;  the  number  2  being  3  times 
contained  in  6,  or  the  third  part  of  it.  In  like  man- 
ner, the  ratio  of  the  quantity,  3  to  6,  is  measured  by 
£  or  | ;  the  ratio  of  4  to  6  is  |  or  § ;  that  of  6  to  4  is 
|  or  |,  &c.  The  ratio  of  two  lines  is  the  ratio  of 
the  number  of  times  which  each  contains  the  common 
measure  of  the  two  lines.  When  the  terms  of  a  ratio 
are  equal,  it  is  called  a  ratio  of  Equality.  When  un- 
equal, a  ratio  of  Inequality. 

76.  Proportion  is  an  equality  of  ratios.     Thus, 

77.  Three  quantities  are  said  to  be  proportional 
when  the  ratio  of  the  first  to  the  second  is  equal  to 
the  ratio  of  the  second  to  the  third.  As  of  the  three 
quantities,  A  =  2,  B  =  4,  C  =  8,  where  f  =  £  =  j,  both 
the  same  ratio. 

78.  Four  quantities  are  said  to  be  proportional 
when  the  ratio  of  the  first  to  the  second  is  the  same 
as  the  ratio  of  the  third  to  the  fourth.  As  of  the  four, 
A  (4),  B  (2),  C  (10),  D  (5),  where  j  =  y  -  2,  both  the 
same  ratio. 

#  The  antecedent  and  consequent  are  called  the  terras  of  a  ratio. 


DEFINITIONS.  51 

Note. — To  denote  that  four  quantities,  A,  B,  (?,  D, 
are  proportional,  they  are  usually  stated  or  placed 
thus,  A  :  B  : :  C  :  D  ;  and  read  thus,  A  is  to  B  as  C  is  to 
D.  The  two  dots  :  must  be  understood  as  represent- 
ing the  sign  of  division  ;  the  four  dots  : :  the  sign  of 
equality.     The  same  proportion  or  equality  of  ratios 

A      C 

may  be  written  thus,  —  =  — ,  or  A  :  B  =  C  :  D.    When 
B      D 

three  quantities  are  proportional,  the  middle  one  is 

repeated,  and  they  are  written  thus,  A  :  B  : :  B :  C. 

79.  Of  three  proportional  quantities,  the  middle  one 
is  said  to  be  a  Mean  Proportional  between  the  other 
two ;  and  the  last  a  Third  Proportional  to  the  first 
and  second. 

80.  Of  four  proportional  quantities,  the  last  is  said 
to  be  a  Fourth  Proportional  to  the  other  three,  taken 
in  order. 

81.  Quantities  are  said  to  be  Continually  Propor- 
tional, or  in  Continued  Proportion,  when  the  ratio  is 
the  same  between  every  two  adjacent  terms,  viz., 
when  the  first  is  to  the  second  as  the  second  to  the 
third,  as  the  third  to  the  fourth,  as  the  fourth  to  the 
fifth,  and  so  on,  all  in  the  same  common  ratio. 

As  in  the  quantities  1,  2,  4,  8,  16,  &c,  where  the 
common  ratio  is  equal  to  2. 

82.  Of  any  number  of  quantities,  A,  B,  C,  D,  the 
ratio  of  the  first  A,  to  the  last  D,  is  said  to  be  com- 
pounded of  the  ratios  of  the  first  to  the  second,  of  the 
second  to  the  third,  and  so  on  to  the  last. 

83.  Inverse  ratio  is,  where  the  antecedent  is  made 
the  consequent,  and  the  consequent  the  antecedent. 
Thus,  if  1  :  2  : :  3  :  6  ;  then  inversely,  or  by  inversion, 
2  :  1  : :  6  :  3. 

84.  Alternate  proportion  is  where  antecedent  is 
compared  with  antecedent,  and  consequent  with  con- 
sequent. As,  if  1  :  2  : :  3  :  6  ;  then,  by  alternation  or 
permutation,  it  will  be  1  :  3  : :  2  :  6. 

85.  Compound  ratio  is,  where  the  sum  of  the  ante- 
cedent and  consequent  is  compared  either  with  the 
consequent  or  with  the  antecedent.     Thus,  if  1:2:: 


52  GEOMETRY. 

3:6;  then,  by  composition,  1  -f-  2  :  1  : :  3  +  G  :  3,  and 
l+2:2::3  +  6:6. 

86.  Divided  ratio  is,  when  the  difference  of  the  an- 
tecedent and  consequent  is  compared  either  with  the 
antecedent  or  with  the  consequent.  Thus,  if  1  :  2  : : 
3:6;  then,  by  division,  2  —  1  :  1  : :  6  —  3  :  3,  and  2  —  1 : 
2::6  — 3:6. 

Note. — The  term  Division  here  means  subtracting, 
or  parting ;  being  used  in  the  sense  opposed  to  com- 
pounding, or  adding,  in  def.  85. 

THEOREM    XLIV. 

Equimultiples  of  any  two  quantities  have  the  same 
ratio  as  the  quantity  themselves. 

Let  A  and  B  be  any  two  quantities,  and  mA,  wzB 
any  equimultiples  of  them,  m  being  any  number  what- 
ever ;  then  will  mA  and  raB  have  the  same  ratio  as  A 
and  B,  or  A  :  B  : :  mA :  wB. 

For  ^  =  |.    Q.E.D. 

mA     A 

Corol.  Hence  like  parts  of  quantities  have  the  same 
ratio  as  the  wholes  ;  because  the  wholes  are  equimul- 
tiples of  the  like  parts,  or  A  and  B  are  like  parts  of 

mA  and  mB. 

■p 

Corol.  2.  If  —  represent  the  first  ratio  of  a  pro- 
portion, or  equality  of  ratios,  — -  must  be  the  form  of 

mA 

the  second  ratio,  m  being  a  quantity  entire  or  frac- 
tional, rational  or  irrational.  In  the  following  theo- 
rems the  form  of  a  proportion  will  always  be  assumed 
in  accordance  with  this  corollary. 

THEOREM    XLV. 

If  four  quantities  of  the  same  kind  be  proportionals, 
they  will  be  in  proportion  by  alternation  or  permutation, 
or  the  antecedents  will  have  the  same  ratio  as  the  conse- 
quents. 


THEOREMS.  53 

Let  A  :  B  : :  mA  :  mB  ;  then  will  A  :  mA  : :  B  :  mB. 

^      mA      m        ,  mB      m   .     .     . 
For  — —  =  — ,  and  -^-  =  — ,  both  the  same  ratio. 
A        1  B        1 


THEOREM    XLVI. 

If  four  quantities  be  proportional,  they  will  be  in 
proportion  by  inversion  or  inversely. 

Let  A  :  B  : :  mA  :  mB  ;  then  will  B  :  A  : :  mB  :  mA. 
-r,  mA     A 

For  ^irs- 

Otherwise.     Let  A  :  B  : :  C  :  D  ;  then  shall  B  :  A  : : 
D:C. 

A        C* 

For  let  —  =  =-=  r ;  then  A  =  Br,  and  C  =  Dr :  there- 
fore B  =  —  and  D  =  — .    Hence  —  =  -,  and  —  =  -. 
r  r  A      r  0      r 

B         T) 

Whence  it  is  evident  that  —  =  —  (ax.  1),  or  B  :  A  : : 

A      O 

D:C. 

In  a  similar  manner  may  most  of  the  other  theo- 
rems of  proportion  be  demonstrated. 

THEOREM    XLVII. 

If  four  quantities  be  proportional,  they  will  be  in 
proportion  by  composition  and  division. 

Let  A  :  B  :  :  mA  :  mB  ; 

Then  will  B  ±  A  :  A  : :  mB  =fc  mA  :  mA, 
and                 B  ±  A  :  B  : :  »iB  ±  mA  :  mB. 
~           mA  A  ,        mB  B 

ror  , s  =7t— ! — ri  anc* 


mBimA     B  =fc  A'         mB  ±  mA     B±A' 
Corol.  1.    If  A  :  B  :  :  mA  :  mB,  then  B  +  A  :  B  — 

A  : :  mB  -f  mA  :  mB  — mA. 

Corol.  2.  It  appears  from  hence  that  the  sum  of  the 

greatest  and  least  of  four  proportional  quantities  ol 

the   same  kind  exceeds  the  sum  of  the  other  two. 

For  since  A  :  A  +  B  :  :  mA  : :  mA  +  mB,  where  A  i? 

the  least,  and  mA  +  mB  the  greatest ;  then  m  +  1 .  A 


54  GEOMETRY. 

+  mB,  the  sum  of  the  greatest  and  least,  evidently 
exceeds  m  +  1  .  A  +  B,  the  sum  of  the  two  other 
quantities,  since  mB  >  B. 

THEOREM    XLVIII. 

If  of  four  proportional  quantities  there  be  taken 
any  equimultiples  whatever  of  the  two  antecedents,  and 
any  equimultiples  whatever  of  the  two  consequents,  the 
quantities  resulting  will  still  be  proportional. 

Let  A  :  B  : :  mA  :  mB  ;  also,  let  pA  andpmA  be  any 
equimultiples  of  the  two  antecedents,  and  qB  and 
qmB  any  equimultiples  of  the  two  consequents ;  then 
will pA  :  qB  :  :pmA  :  qmB. 

For  gmB_qB^ 

pmA    pA' 

THEOREM    XLIX. 

If  there  be  four  proportional  quantities,  and  the  two 
consequents  be  either  augmented  or  diminished  by  quan- 
tities that  have  the  same  ratio  as  the  respective  antece- 
dents, the  results  and  the  antecedents  will  still  be  pro- 
portionals. 

Let  A  :  B  :  :  mA  :  mB,  and  nA  and  nmA  any  two 
quantities  having  the  same  ratio  as  the  two  antece- 
dents ;  then  will  A  :  B  =t  nA  :  :  mA  :  mB  ±  nmA. 

-r,  mB  rb  n?nA      B  =t  nA 

For  = . 

mA  A 

THEOREM    L. 

If  any  number  of  quantities  be  proportional,  then 
any  one  of  the  antecedents  will  be  to  its  consequent  as 
the  sum  of  all  the  antecedents  is  to  the  sum  of  all  the 
consequents. 

Let  A  :  B  : :  mA  :  mB  : :  nA  :  nB,  &c. ;  then  will  A  : 
B  : :  A  +  mA  +  nA  :  B  -f  mB  -f  nB,  &c. 

B  +  mB  +  nB      (1  +  m  +  n)  B     B 


For 


A  +  mA  -{-nA     ( 1  -f-  m  -|-  »)  A     A 


THEOREMS.  55 


THEOREM    LI. 

If  a  whole  magnitude  be  to  a  whole  as  a  part  taken 
from  the  first  is  to  a  part  taken  from  the  other,  then 
the  remainder  will  be  to  the  remainder  as  the  whole  to 
the  whole. 


Let 

A        T>           m     A       mT> 

A  :  B  : :  —  A  :  — B; 

n         n 

then  will 

A:B::A-™A:B_™B. 

n                 n 

For 

B--B      R 
n         B 

a-^a"a' 

THEOREM    LII. 

If  any  quantities  be  proportional,  their  squares  or 
cubes,  or  any  like  powers  or  roots  of  them,  will  also 
be  proportional. 

Let  A  :  B  : :  mA  :  mB  ;  then  will  A" :  Bn : :  mnAn  : 

mnBn. 

„  m"Bn      B" 

For  -^T^=  TV 

m  A"      A 

THEOREM     LIII. 

If  there  be  two  sets  of  proportionals,  then  the  prod- 
ucts or  rectangles  of  the  corresponding  terms  will  also 
be  proportional. 

Let  A  :  B  : :  mA  :  mB, 

and  C  :  D  : :  wC  :  wD  ; 

then-will        AC  :  BD  : :  mnAC  :  mnBD. 

„  mnBD     BD 

For =  — . 

m?iAC     AC 


56  GEOMETRY. 


THEOREM    L1V. 

If  four  quantities  be  proportional,  the  rectangle  or 
product  of  the  two  extremes  will  be  equal  to  the  rect- 
angle or  product  of  the  two  means.     And  the  converse. 

Let  A  :  B  : :  mA  :  mB  ; 

then  is  A  X  mB  =  B  X  mA  =  mAB,  as  is  evident. 

THEOREM    LV. 

If  three  quantities  be  continued  proportionals,  the 
rectangle  or  product  of  the  two  extremes  will  be  equal 
to  the  square  of  the  mean.     And  the  converse. 

Let      A,  mA,  maA  be  three  proportionals, 
or  A  :  mA  : :  mA  :  m*A  ; 

then  is       A  X  m2A  =  m2Aa,  as  is  evident. 

* 

THEOREM    LVI. 

If  any  number  of  quantities  be  continued  propor- 
tionals, the  ratio  of  the  first  to  the  third  will  be  du- 
plicate, or  the  square  of  the  ratio  of  the  first  and  second; 
and  the  ratio  of  the  first  and  fourth  will  be  triplicate, 
or  the  cube  of  that  of  the  first  and  second,  and  so  on. 

Let   A,  mA,  rn^A,  m*A,  &c,  be  proportionals  ; 

then  is— -  =  — ;  but— -  =  —  ;  and-— =  — ,  &c. 
mA      m  mA     m  mA     m 

THEOREM    LVII. 

Triangles,  and  also  parallelograms,  having  equal 
altitudes,  are  to  each  other  as  their  bases. 

Let  the  two  triangles  ADC,  DEF  have  x  ck  F 
the  same  altitude,  or  be  between  the  same 
parallels  AE,  IF ;  then  is  the  surface  of 
the  triangle  ADC  to  the  surface  of  the 
triangle  DEF  as  the  base  AD  is  to  the 
base  DE.  Or  AD  :  DE  : :  the  triangle  A  B  D  G  H  E 
ADC  :  the  triangle  DEF. 


i 


THEOREMS.  57 

For,  let  the  base  AD  be  to  the  base  DE  as  any  one 
number  m  (which  we  have  taken  2  in  the  diagram), 
to  any  other  number  n  (which  we  have  taken  3  in  the 
diagram,  though  the  reasoning  would  be  the  same  for 
any  other  numbers) ;  and  divide  the  respective  bases 
into  those  parts,  AB,  BD,  DG,  GH,  HE,  all  equal  to 
one  another ;  and  from  the  points  of  division  draw  the 
lines  BC,  GF,  HF  to  the  vertices  C  and  F.  Then 
will  these  lines  divide  the  triangles  ADC,  DEF  into 
the  same  number  of  parts  as  their  bases,  each  equal 
to  the  triangle  ABC,  because  those  triangular  parts 
have  equal  bases  and  altitudes  (cor.  2,  th.  22) ;  name- 
ly, the  triangle  ABC  equal  to  each  of  the  triangles 
BDC,  DFG,  GFH,  HFE.  So  that  the  triangle  ADC 
is  to  the  triangle  DFE  as  the  number  of  parts  m  (2) 
of  the  former  to  the  number  n  (3)  of  the  latter,  that  is, 
as  the  base  AD  to  the  base  DE.  (See  end  of  note  to 
def.  75.) 

The  parallelograms  ADKI,  DEFK  being  doubles 
of  the  triangles  ACD,  DFE,  are  in  the  same  ratio, 
viz.,  that  of  the  base  AD  to  the  base  DE.     Q.  E.  D 

THEOREM    LVIII. 

Triangles,  and  also  parallelograms,  having  equal 
bases,  are  to  each  other  as  their  altitudes. 

Let  ABC,  BEF  be  two  triangles 
having  the  equal  bases  AB,  BE,  and 
whose  altitudes  are  the  perpendicu- 
lars CG,  FH  ;  then  will  the  triangle 
ABC  :  the  triangle  BEF  : :  CG  :  FH. 

For,  let  BK  be  perpendicular  to  A  G  B  H  E 
AB,  and  equal  to  CG ;  in  which  let  there  be  taken 
BL  =  FH  ;  drawing  AK  and  AL. 

Then,  triangles  of  equal  bases  and  heights  being 
equal  (cor.  2,  th.  22),  the  triangle  ABK  is  =  ABC,  nnd 
the  triangle  ABL  =  BEF.  The  two  triangles  ABK, 
ABL  may,  therefore,  be  compared,  instead  of  the  two 
given  triangles,  which  are  respectively  equal  to  them ; 
and  having  the  same  altitude  AB,  they  will  be  as 
C2 


58  GEOMETRY. 

their  bases  (th.  57),  namely,  the  triangle  ABK  :  the 
triangle  ABL  : :  BK  =  CG  :  BL  =  FH. 

Therefore,  the  triangle  ABC  :  triangle  BEF  : :  CG 
:FH. 

And  since  parallelograms  are  the  doubles  of  these 
triangles,  having  the  same  bases  and  altitudes,  they 
will  likewise  have  to  each  other  the  same  ratio  as 
their  altitudes.     Q.  E.  D. 

THEOREM    LIX. 

If  four  lines  be  proportional,  the  rectangle  of  the 
extremes  will  be  equal  to  the  rectangle  of  the  means; 
and,  conversely,  if  the  rectangle  of  the  extremes  of  four 
lines  be  equal  to  the  rectangle  of  the  means,  the  four 
lines  will  be  proportional. 

Let  the  four  lines  A,  B,  C,  D  be 
proportionals,  or  A  :  B  : :  C  :  D  ;  then 
will  the  rectangle  of  A  and  D  be 
equal  to  the  rectangle  of  B  and  C ; 
or  the  rectangle  A  .  D  =  B  .  C. 

For,  let  the  four  lines  be  placed 
with  their  four  extremities  meeting  in  a  common  point, 
forming  at  that  point  four  right  angles ;  and  draw 
lines  parallel  to  them  to  complete  the  rectangles  P, 
Q,  R,  where  P  is  the  rectangle  of  A  and  D,  Q  the 
rectangle  of  B  and  C,  and  R  the  rectangle  of  B  and  D. 

Then  the  rectangles  P  and  R,  being  between  the 
same  parallels,  are  to  each  other  as  their  bases  A  and 
B  (th.  57) ;  and  the  rectangles  Q,  and  R,  being  be- 
tween the  same  parallels,  are  to  each  other  as  their 
bases  C  and  D.  But  the  ratio  of  A  to  B  is  the  same 
as  the  ratio  of  C  to  D,  by  hypothesis ;  therefore,  the 
ratio  of  P  to  R  is  the  same  as  the  ratio  of  Q  to  R, 
and,  consequently,  the  rectangles  P  and  Q,  are  equal. 
Q.  E.  D. 

Again,  if  the  rectangle  of  A  and  D  be  equal  to  the 
rectangle  of  B  and  C,  these  lines  will  be  proportional, 
or  A  :  B  : :  C  :  D. 

For,  the  rectangles  being  placed  the  same  as  be- 


QC 
B 

A 

R 


D      P 

THEOREMS.  59 

fore  ;  then,  because  parallelograms  between  the  same 
parallels  are  to  one  another  as  their  bases,  the  rect- 
angle P  :  R  :  :  A  :  B,  and  Q  :  R  : :  C  :  D.  But  as  P 
and  Q,  are  equal  by  supposition,  they  have  the  same 
ratio  to  R,  that  is,  the  ratio  of  A  to  B  is  equal  to  the 
ratio  of  C  to  D,  or  A  :  B  : :  C  :  D.     Q.  E.  D. 

Corol.  1.  When  the  two  means,  namely,  the  second 
and  third  terms,  are  equal,  their  rectangle  becomes  a 
square  of  the  second  term,  which  supplies  the  place 
of  both  the  second  and  the  third.  And  hence  it  fol- 
lows that,  when  three  lines  are  proportionals,  the  rect- 
angle of  the  two  extremes  is  equal  to  the  square  of 
the  mean ;  and,  conversely,  if  the  rectangle  of  the 
extremes  be  equal  to  the  square  of  the  mean,  the  three 
lines  are  proportionals. 

Corol.  2.  If  the  sides  about  the  equal  angles  of  par- 
allelograms or  triangles  be  reciprocally  proportional,*' 
the  parallelograms  or  triangles  will  be  equal ;  and, 
conversely,  if  the  parallelograms  or  triangles  be  equal, 
their  sides  about  the  equal  angles  will  be  reciprocally 
proportional.  It  is  only  necessary  to  suppose  P  and 
Q  parallelograms  to  prove  this.     (See  also  th.  19.) 

THEOREM    LX. 

Rectangles  are  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 

For,  in  the  last  figure,  let  the  two  rectangles  P  and 
Q  be  unequal,  and  be  placed  as  before.    Then  (th.  57), 
P:R::  A:B; 
R  :  Q  : :  D  :  C. 
Multiplying  the  two  proportions,  and  striking  out 
the  common  factor  R  from  the  two  terms  of  the  first 
ratio  of  the  resulting  proportion,  we  have 
P:Q::  AxD:BxC.     Q.  E.  D. 
Scholium.  The  area  or  space  of  a  rectangle  may, 

*  i.  c,  the  first  side  of  the  first  is  to  the  first  of  the  second  as  the 
second  of  the  second  is  to  tlit*  second  of  the  first.  By  multiplying  the 
extremes  and  means,  this  will  make  the  rectangle  of  the  two  sides  of 
the  one  figure  equal  to  the  rectangle  of  the  two  sides  of  the  other. 


CO  GEOAJETKY. 

then,  be  represented  or  expressed  by  the  product  of 
its  length  and  breadth  multiplied  together.  And,  in 
geometry,  the  rectangle  of  two  lines  signifies  the  same 
thing  as  their  product.  (Compare  ths.  59,  54.)  Also, 
a  square  is  similar  to,  or  represented  by,  its  side  mul- 
tiplied by  itself,  or  written  with  an  exponent  2. 

CoroL  1.  Since,  by  th.  22,  corol.  2,  rhomboids  are 
equivalent  to  rectangles  having  the  same  base  and 
altitude,  it  follows  that  the  areas  of  all  parallelograms 
will  be  expressed  by  the  product  of  the  base  by  the 
altitude,  and  of  triangles  which  are  the  halves  of 
parallelograms  of  the  same  base  and  altitude,  by  the 
product  of  the  base  by  half  the  altitude,  or  the  altitude 
by  half  the  base,  or  half  the  product  of  the  base  by 
the  altitude. 

Corol.  2.  Parallelograms  or  triangles  having  equal 
bases  will  be  to  each  other  as  their  altitudes ;  those 
having  equal  altitudes  will  be  to  each  other  as  their 
bases ;  and  those  having  neither  equal  will  be  as  the 
products  of  their  bases  by  their  altitudes. 

Corol.  3.  Parallelograms  or  triangles  having  an 
angle  in  each  equal,  are  in  proportion  to  each  other 
as  the  rectangles  of  the  sides  which  are  about  these 
equal  angles.  This  may  be  proved  from  the  last  dia- 
gram, supposing  P  and  Q  to  be  parallelograms. 

THEOREM    LXI. 

If  a  line  be  drawn  in  a  triangle  parallel  to  one  of  its 
sides,  it  will  cut  the  other  two  sides  proportionally. 

Let  DE  be  parallel  to  the  side  BC 
of  the  triangle  ABC ;  then  will  AD  : 
DB  : :  AE  :  EC. 

For,  draw  BE  and  CD.  Then  the 
triangles  DBE,  DCE  are  equal  to  each 
other,  because  they  have  the  same  base 
DE,  and  are  between  the  same  paral- 
lels DE,  BC  (th.  22).  But  the  two  tri-  B  c 
angles  ADE,  BDE,  on  the  bases  AD,  DB,  have  the 
same  altitude,  viz.,  the  perpendicular  from  their  com- 


THEOREMS.  01 

mon  vertex  E  to  the  line  of  their  bases  BA ;  and  the 
two  triangles  ADE,  CDE,  on  the  bases  AE,  EC,  have 
also  a  common  altitude  ;  and  because  triangles  of  the 
same  altitude  are  to  each  other  as  their  bases,  there- 
fore 

the  triangle      ADE  :  BDE  : :  AD  :  DB, 
and  triangle     ADE  :  CDE  : :  AE  :  EC. 

But  BDE  is  =  CDE ;  hence  the  first  ratio  is  the 
same  in  these  two  proportions,  and  the  second  ratios 
must  be  equal ;  therefore  AD  :  DB  : :  AE  :  EC.  Q. 
E.  D. 

CoroL  1.  Also,  the  whole  lines  AB,  AC  are  propor- 
tional to  their  corresponding  proportional  segments 
(th.  47), 

viz.,  AB  :  AC  : :  AD  :  AE, 

and  AB  : AC : : BD :  CE. 

CoroL  2.  The  converse  of  the  above  proposition  is 
also  true,  viz.,  that  a  line  which  divides  the  two  sides 
of  a  triangle  proportionally  must  be  parallel  to  the 
base.  For  any  other  line  through  D  than  the  parallel 
DE,  /neeting  AC  in  some  other  point  than  E,  must 
divide  AC  into  two  parts,  having  a  different  ratio 
from  AE  to  EC,  and,  consequently,  different  from  the 
ratio  AD :  DB. 

THEOREM    LXII. 

A  line  which  bisects  any  angle  of  a  triangle  divides 
the  opposite  side  into  two  segments,  which  are  propor- 
tional to  the  other  two  adjacent  sides. 

Let  the  angle  BAC,  of  the  tri-    E 
angle  ABC,  be  bisected  by  the     \\ 
line  AD  ;  then  will  the  segment      \    ***-j^ 
BD  be  to  the  segment  DC  as 
the  side  AB  is  to  the  side  AC. 

For,  let  BE  be  drawn  parallel 
to  AD,  meeting  C A  produced  at  B       D  C 

E.  Then,  because  the  line  BA  meets  the  two  paral- 
lels AD,  BE,  it  makes  the  angle  ABE  equal  to  the  al- 
ternate angle  s.  (th.  10),  and  therefore  also  equal  to 


62  GEOMETRY. 

the  angle  r,  which  is  (by  hyp.)  equal  to  s.  Again, 
because  the  line  CE  cuts  the  two  parallels  AD,  BE,  it 
makes  the  angle  E  equal  to  the  angle  r*  on  the  same 
side  (th.  10).  Hence,  in  the  triangle  ABE,  the  an- 
gles B  and  E,  being  each  equal  to  half  the  bisected 
angle  of  the  triangle,  are  equal  to  each  other,  and, 
consequently,  their  opposite  sides  AB,  AE  are  also 
equal  (th.  4). 

But  now,  in  the  triangle  CBE,  the  line  AD,  being 
parallel  to  the  side  BE,  cuts  the  other  two  sides,  CB, 
CE,  proportionally  (th.  61),  making  CD  to  DB,  as  is 
CA  to  AE,  or  to  its  equal  AB.     Q.  E.  D. 

THEOREM    LXIII. 

Equiangular  triangles  are  similar,  or  have  their 
like  sides  proportional. 

For,  by  th.  43,  the  rectangles  of  the  corresponding 
sides  taken  alternately  are  equal,  and  by  the  second 
part  of  th.  59,  these  corresponding  or  like  sidesf  are, 
in  consequence,  directly  proportional. 

THEOREM    LXIV. 

Triangles  which  have  their  sides  proportional  are 
also  equiangular. 

In  the  two   triangles   ABC,  DEF,  if  c 

AB  :  DE  : :  AC  :  DF  : :  BC  :  EF,  the  two 

triangles  will  have  their  corresponding 

angles  equal.  a  b 

For,  if  the  triangle  ABC  be  not  equian-  G    F 

gular  with   the  triangle  DEF,  suppose  ,  \  "\ 

some  other  triangle,  as  DEG,  constructed       //  \\ 
upon  the  side  DE,  to  be  equiangular  with     -^  \ 

ABC.     But  this  is  impossible  ;  for  if  the  D  E 

two  triangles  ABC,  DEG  were  equiangular,  their 
sides  would  be  proportional  (th.  63),  viz., 

*  The  use  of  small  letters  to  designate  angles  may  be  adopted  in 
other  propositions. 

t  The  corresponding  sides  are  called  homologous- 


THEOREMS.  6& 

AB  :  DE  : :  AC  :  DG, 

but,  by  hypothesis, 

AB  :  DE  : :  AC  :  DF  ; 

.-.  DG  -  DF. 

In  the  same  manner,  it  may  be  proved  that 

EG  -  EF  ; 

.-.  (th.  5),  A*  DEF  is  identical  with  A  DEG,  which  is 

absurd,  the  angles  being  different. 

TIIEOBEM    LXV. 

Triangles  which  have  an  angle  in  the  one  equal  to 
an  angle  in  the  other,  and  the  sides  about  these  angles 
proportional,  are  equiangular. 

Let  ABC,  DEF  be  two      A 

triangles,  having  the  angle        /\ 

A  =  the  angle  D,  and  the       /   \  D 

sides  AB,  AC  proportional      /       \  \ 

to  the  sides  DE,  DF  ;  then   (d AH  I  \ 

will   the   triangle  ABC   be     /  \  /     \ 

equiangular  with  the  trian- 
gle DEF. 

For,  make  AG  -  DE,  and  AH  =  DF,  and  join  GH. 

Then  the  two  triangles  DEF,  AGH,  having  two 
sides  equal,  and  the  contained  angles  A  and  D  equal, 
are  identical  and  equiangular  (th.  1),  having  the  an- 
gles G  and  H  equal  to  the  angles  E  and  F.  But, 
since  the  sides  AG,  AH  are  proportional  to  the  sides 
AB,  AC,  the  line  GH  is  parallel  to  BC  (th.  61,  corol. 
2) ;  hence  the  angles  B  and  C  are  equal  to  the  angles 
G  and  H  respectively  (th.  10),  and,  consequently,  to 
their  equals  E  and  F.     Q.  E.  D. 

General  Scholium. — Triangles  will  be  similar,  1°. 
When  they  have  their  angles  equal,  or  two  of  their 
angles  equal  (th.  15,  corol.  1) ;  2°.  When  they  have 
their    homologous    sides    proportional  ;f    3°.  When 

*  This  sign  (A)  stands  for  the  word  triangle. 

\  Triangles  are  the  only  polygons  in  which  one  part  of  the  defini- 
tion (def.  07  )  of  similar  figures  involves  the  other  as  a  necessary  con- 
sequence.    Thus  a  square  and  a  rectangle  are  equiangular  quadri- 


B  C     E 


64  GEOMETRY. 

they  have  an  angle  in  each  equal,  and  the  sides  about 
the  equal  angles  proportional  ;  4°.  When  they  have 
their  sides  respectively  parallel  or  perpendicular,  or 
in  any  way  equally  inclined. 

THEOREM    LXVI. 

In  a  right-angled  triangle,  a  peipendicular  from 
the  right  angle  is  a  mean  proportional  between  the 
segments  of  the  hypothenuse,  and  each  of  the  sides 
about  the  right  angle  is  a  mean  proportional  between 
the  hypothenuse  and  the  adjacent  segment. 

Let  ABC  be  a  right-angled  tri- 
angle, and  AD  a  perpendicular  from 
the  vertex  of  the  right  angle  A  to 


the  hypothenuse  CB ;  then  will  B  DC 

AD  be  a  mean  proportional  between  BD  and  DC  ; 
AB  a  mean  proportional  between  BC  and  BD  : 
AC  a  mean  proportional  between  BC  and  DC. 

For,  the  two  right-angled  triangles  ABD,  ABC, 
having  the  angle  B  common,  are  equiangular  (cor.  2, 
th.  15).  For  a  similar  reason,  the  two  triangles  ABC, 
ADC  are  equiangular. 

Hence,  then,  all  the  three  triangles,  viz.,  the  whole 
triangle   and  the  two  partial   triangles  ABC,  ABD, 
ADC,  being  equiangular,  will  have  their  like  sides 
proportional  (th.  63), 
viz.,*  BD:  AD::  AD:  DC; 

and  BC  :  AC  : :  AC  :  DC  ; 

and  BC  :  AB  : :  AB  :  BD.  Q.  E.  D. 

laterals,  but  their  homologous  sides  are  not  proportional,  the  adjacent 
ones  of  the  square  having  a  ratio  of  equality,  those  of  the  rectangle 
a  ratio  of  inequality.  Again,  the  sides  of  a  square  and  rhombus  are 
proportional,  having  in  both  the  ratio  of  equality,  but  the  angles  are 
not  equal,  those  oi  the  square  being  right,  those  of  the  rhombus 
oblique. 

*  The  student  will  be  aided  by  saying  BD,  the  long  perpendicular 
side  of  the  left  triangle,  is  to  AD,  the  long  perpendicular  side  of  the 
right  triangle,  as  AD,  the  short  perpendicular  side  of  the  former,  is 
to  DC,  the  short  perpendicular  side  of  the  latter.  Again,  BC,  the  hy- 
pothenuse of  the  whole  triangle,  is  to  AB,  the  hypothenuse  of  the 
left  partial  triangle,  &c. 


THEOREMS.  65 

Corol.  1.  Because  the  angle  in  a  semicircle  is  a 
right  angle  (corol.  3,  th.  39),  it  follows  that  if,  from 
any  point  A  in  the  periphery  of  the  semicircle,  a  per- 
pendicular be  drawn  to  the  diameter  BC,  and  the 
two  chords  CA,  AB  be  drawn  to  the  extremities  of 
the  diameter ;  then  are  AD,  AB,  AC  the  mean  pro- 
portionals as  in  this  theorem,  or  (by  th.  55)  AD2  = 
CD  .  BD  ;  ABa  =  BC  .  BD ;  and  AC3  =  CB  .  CD. 

Corol  2.  Hence  AB2 :  AC2 : :  CD  :  BD. 

Corol.  3.  Hence  we  have  another  demonstration 
of  th.  26. 

For,  since     AB2  =  BC  .  BD,  and  AC2  =  BC . CD  ; 
By  addition,  AB2  +  AC2  =  BC  (BD  +  CD)  =  BC2. 

THEOREM  LXVII. 

Similar  triangles  are  to  each  other  as  the  squares  of 
their  like  sides. 

Let  ABC,  DEF  be  two 
similar  triangles,  AB  and 
DE  being  two  like  sides ; 
then  will  the  triangle  ABC 
be  to  the  triangle  DEF  as  gj~- 
the  square  of  AB  is  to  the 
square  of  DE,  or  as  AB2  to 
DE*.  B  C    E  ± 

For,  the  triangles  being  similar,  they  have  their 
like  sides  proportional  (def.  67)  ; 
therefore  AB  :  DE  : :  AC  :  DF  ; 

and     AB  :  DE  : :  AB  :  DE,  an  identity  of  ratios  ; 
therefore  AB2 :  DE2 : :  AB .  AC  :  DE  .  DF  (th.  53). 

But  the  triangles  are  to  each  other  as  the  rectangles 
of  the  like  pairs  of  their  sides  (cor.  3,  th.  60)  ;  or 

A  ABC  :  A  DEF : :  AB .  AC  :  DE  .  DF  ; 
therefore      A  ABC  :  A  DEF : :  AB2 :  DE2.       Q.  E.  D. 

THEOREM   LXVIII. 

The  perimeters  of  all  similar  figures  are  to  each 
other  as  their  homologous  sides,  and  the  surfaces  as  the 
squares  of  their  homologous  sides. 


66 


GE0MET11Y. 


Let    ABCDE, 
FGHIK  be  any 

two  similar  fig- 
ures, their  like 
sides  being  AB, 
FG,andBC,GH, 
and  so  on  in  the 
same  order ;  then 
will  the  perimeter  of  the  figure  ABCDE  be  to  the  pe- 
rimeter of  the  figure  FGHIK  as  AB  to  FG,  and  the 
surface  as  the  square  of  AB  to  the  square  of  FG,  or 
as  AB2  to  FG2. 

For  (by  def.  67)  AB  :  BC  :  CD,  &c.  :  :  FG  :  GH  : 
HI,  &c.  And  (by  th.  50)  AB  +  BC  +  CD,  &c. ;  or 
the  perimeter  of  the  first  polygon  is  to  FG  -f-  GH  + 
HI,  &c. ;  or  the  perimeter  of  the  second  polygon  as 
AB  :  FG. 

Again,  draw  AC,  AD,  FH,  FI,  dividing  the  figures 
into  an  equal  number  of  triangles  by  lines  from  two 
equal  angles  A  and  F. 

The  two  figures  being  similar  (by  hyp.),  they  are 
equiangular,  and  have  their  like  sides  proportional 
(def.  67). 

Then,  since  the  angle  B  is  =  the  angle  G,  and  the 
sides  AB,  BC  proportional  to  the  sides  FG,  GH,  the 
triangles  ABC,  FGH  are  equiangular  (th.  65).  If, 
from  the  equal  angles  BCD,  GHI  there  be  taken  the 
equal  angles  ACB,  GHF,  there  will  remain  the  equals 
ACD,  FHI ;  and  since,  from  the  similarity  of  the  tri- 
angles ABC,  FGH,  and  of  the  whole  polygons,  AC 
and  FH,  as  well  as  CD  and  HI,  have  the  same  ratio 
that  BC  and  GH  have,  they  must  have  the  same  ratio 
as  one  another ;  hence  the  triangles  ACD,  FHI,  hav- 
ing an  equal  angle  contained  by  proportional  sides 
are  (th.  65)  similar. 

In  the  same  manner,  ADE  may  be  proved  similar 
to  FIK.  Hence  each  triangle  of  the  one  figure  is  equi- 
angular with  each  corresponding  triangle  of  the  other. 

But  equiangular  triangles  are  similar  (th.  63),  and 
are  proportional  to  the  squares  of  their  like  sides 
(th.  67). 


THEOREMS. 


67 


B 


Therefore  the  A  ABC  :  A  FGH  : :  AB2 :  FG2 ; 
and  A  ACD  :  A  FHI   : :  DC2 :  HP  ; 

and  A  ADE  :  A  FIK    : :  DE2 :  IK2. 

But  as  the  two  polygons  are  similar,  their  like  sides 
are  proportional,  and,  consequently,  their  squares  also 
roportfonal ;  so  that  all  the  ratios  AB2  to  FG2,  and 
Cu  to  HI2,  and  DEa  to  IK2,  are  equal  among  them- 
selves, and,  consequently,  the  corresponding  triangles 
als...  ABC  to  FGH,  and  ACD  to  FHI,  and  ADE  to 
FIK,  have  all  the  same  ratio,  viz.,  that  of  AB2  to 
FG2 ;  and  hence  the  sum  of  the  antecedents,  or  the 
figure  ABCDE,  have  to  the  sum  of  the  consequents, 
or  the  figure  FGHIK,  still  the  same  ratio,  viz.,  that 
of  AB2  to  FG2  (th.  50).     Q.  E.  D. 


THEOREM    LXIX. 

Similar  figures  inscribed  in  circles  have  their  like 
sides,  and  also  their  whole  perimeters,  in  the  same  ratio 
as  the  diameters  of  the  circles  in  which  they  are  in- 
scribed. 

Let  ABCDE,  FGH 

IK  be  two  similar  fig- 
ures, inscribed  in  the 
circles  whose  diame- 
ters are  AL  and  FM ; 
then  will  each  side  AB, 
BC,  &c,  of  the  one 
figure  be  to  the  like  side  FG,  GH,  &c,  of  the  other 
figure,  or  the  whole  perimeter  AB  +  BC  +,  &c,  of 
the  one  figure,  to  the  whole  perimeter  FG  -f-  GH  +, 
&c,  of  the  other  figure,  as  the  diameter  AL  to  the 
diameter  FM. 

For,  draw  the  two  corresponding  diagonals,  AC, 
FH,  as  also  the  lines  BL,  GM.  Then,  since  the  pol- 
ygons are  similar,  they  are  equiangular,  and  their 
like  sides  have  the  same  ratio  (def.  67) ;  therefore  the 
two  triangles  ABC,  FGH  have  the  angle  B  =  the  an- 
gle G,  and  the  two  sides  AB,  BC  proportional  to  the 
two  sides  FG,  GH  ;  consequently,  these  two  triangles 


63 


GEOMETRY. 


are  equiangular  (th.  65),  and  have  the  angle  ACB  = 
FHG.  But  the  angle  ACB  =  ALB,  standing  on  the 
same  arc  AB  ;  and  the  angle  FHG  =  FMG,  stand- 
ing on  the  same  arc  FG ;  therefore  the  angle  ALB 
=  FMG  (ax.  1).  And  since  the  angle  ABL  =  FGM, 
being  both  right  angles,  because  in  a  semicircle ; 
therefore  the  two  triangles  ABL,  FGM,  having  two 
angles  equal,  are  equiangular ;  and,  consequently, 
their  like  sides  are  proportional  (th.  63) ;  hence  AB  : 
FG  : :  the  diameter  AL  :  the  diameter  FM. 

In  like  manner,  each  side  BC,  CD,  &c,  has  to  each 
side  GH,  HI,  &c.,  the  same  ratio  of  AL  to  FM  ;  and, 
consequently,  the  sums  of  them  are  still  in  the  same 
ratio,  viz.,  AB  -f-  BC  +  CD,  &c. :  FG  +  GH  +  HI, 
&c,  :  :  the  diam.  AL  :  the  diam.  FM  (th.  50).  Q. 
E.  D.* 


THEOREM    LXX. 

Similar  figures  inscribed  in  circles  are  to  each  other 
as  the  squares  of  the  diameters  of  those  circles. 

Let  (see  last  fig.)  ABCDE,  FGHIK  be  two  simi- 
lar figures,  inscribed  in  the  circles  whose  diameters 

*  Similar  figures 
admit  of  a  better 
definition  than  that 
given  at  definition 
67,  and  which  can 
now  be  understood. 

They  are  those 
which  can  be  so 
placed  that  lines 
drawn  through  the 
angular  points  of  the 

one  from  some  point  O,  within  or  without, 
shall  also  pass  through  the  angular  points 
of  the  other ;  and  the  distances  from  the 
angular  points  of  the  two  figures  to  the  point 
O  shall  be  proportional. 

This  definition  admits  of  being  enlarged. 
Similar  geometrical  magnitudes  are  those 
which  admit  of  lines  being  drawn  from  a 
point  through  the  corresponding  points  ol 
both,  the  distances  of  which  from  the  radiant  point  are  proportional 
(See  Appendices  II.  and  V.) 


THEOREMS. 


are  AL  and  FM ;  then  the  surface  of  the  pol 
ABCDE  will  be  to  the  surface  of  the  polygon  FGHIK 
as  ALa  to  FAP. 

For  the  figures,  being  similar,  are  to  each  other  as 
the  squares  of  their  like  sides,  ABa  to  FG2  (th.  68). 
But  by  the  last  theorem,  the  sides  AB,  FG  are  as  the 
diameters  AL,  FM  ;  and,  therefore,  the  squares  of  the 
sides  AB'  to  FGa  as  the  squares  of  the  diameters  ALa 
to  FAT  (th.  5*2).  Consequently,  the  polygons  ABCDE, 
FGHIK  are  also  to  each  other  as  the  squares  of  the 
diameters  AL3  to  FM'  (ax.  1).     Q.  E.  D. 

THEOREM    LXXI. 

The  circumferences  of  all  circles  are  to  each  other 
as  their  diameters. 

Let  D,  d  denote  the  diameters  of  two  circles,  and 
C,  c  their  circumferences  ;  then  will  D  :  d : :  C  :  c,  or 
D  :  C  : :  d :  c. 

For  (by  th.  G9)  similar  polygons  inscribed  in  cir- 
cles have  their  perimeters  in  the  same  ratio  as  the 
diameters  of  those  circles. 

Now,  as  this  property  belongs  to  all  polygons, 
whatever  may  be  the  number  of  the  sides,  conceive 
the  number  of  the  sides  to  be  indefinitely  great,  and 
the  length  of  each  infinitely  small.  But  the  perime- 
ter of  the  polygon  of  an  indefinite  number  of  sides 
becomes  the  same  thing  as  the  circumference  of  the 
circle.  Hence  it  appears  that  the  circumferences  of 
circles,  being  the  same  as  the  perimeters  of  such  pol- 
ygons, are  to  each  other  in  the  same  ratio  as  the  di- 
ameters of  the  circles.     Q.  E.  D. 

Corol.  1.  Since  the  radius  is  half  the  diameter, 
circumferences  are  also  as  their  radii. 

Corol.  2.  Similar  arcs  being  the  same  parts  of  their 
respective  circumferences,  are  to  each  other  as  their 
radii. 

Corol.  3.  The  ratio  of  the  arc  which  subtends  an 
angle  to  its  radius,  being  an  arc  which  subtends  the 
same  angle  in  the  circle  whose  radius  is  unity,  it  fol- 


70  GEOMETRY. 

lows  that  angles  at  the  centers  of  different  circles 
are  to  each  other  as  the  ratio  of  the  arcs  which  sub- 
tend them  to  their  radii. 

THEOREM   LXXII. 

The  areas  or  spaces  of  circles  are  to  each  other  as 
the  squares  of  their  diameters,  or  of  their  radii. 

Let  A,  a  denote  the  areas  or  spaces  of  two  circles, 
and  D,  d  their  diameters  ;  then  A  :  a  : :  D? :  d9. 

For  (by  th.  70)  similar  polygons  inscribed  in  cir- 
cles are  to  each  other  as  the  squares  of  the  diameters 
of  the  circles. 

Hence,  conceiving  the  number  of  the  sides  of  the 
polygons  to  be  increased  more  and  more,  or  the 
length  of  the  sides  to  become  less  and  less,  the  pol- 
ygon approaches  nearer  and  nearer  to  the  circle,  till 
at  length,  by  an  infinite  approach,  they  coincide,  and 
become,  in  effect,  equal ;  and  then  it  follows  that  the 
spaces  of  the  circles,  which  are  the  same  as  of  the 
polygons,  will  be  to  each  other  as  the  squares  of  the 
diameters  of  the  circles.     Q.  E.  D. 

Corol.  The  spaces  of  circles  are  also  to  each  other 
as  the  squares  of  the  circumferences  ;  since  the  cir- 
cumferences are  in  the  same  ratio  as  the  diameters 
(by  th.  71). 

THEOREM   LXXIII. 

The  area  of  any  circle  is  equal  to  the  rectangle  of 
half  its  circumference  and  half  its  diameter. 

Conceive  a  regular  polygon  to  v  ^ — ~^E 

be  inscribed  in  a  circle,  and  radii  //\        /  v\ 

drawn  to  all  the  angular  points,  1/     \q/       \\ 

dividing  it  into  as  many  equal  tri-  Ak- yfc yiD- 

angles  as  the  polygon  has  sides,  v\      /|\  H/J 

one  of  which  is  OBC,  of  which  the  \^/    \$\J/ 

altitude  is  the  perpendicular  OG,*  B^ — — ~^C 

*  The  line  OG  is  called  the  apophthegm  of  the  polygon. 


EXERCISES.  71 

from  the  center  to  the  base  BC.  The  other  triangles 
will  have  the  same  altitude  (th.  35). 

Then  the  triangle  OBC  is  equal  (th.  60,  cor.  1)  to 
the  rectangle  of  the  half  base  BC  and  the  altitude 
(  K ;  ;  consequently,  the  whole  polygon,  or  all  the  tri- 
angles added  together  which  compose  it,  is  equal  to 
the  rectangle  of  the  common  altitude  OG,  and  the 
halves  of  all  the  sides,  or  the  half  perimeter  of  the 
polygon. 

Now,  conceive  the  number  of  sides  of  the  polygon 
to  be  indefinitely  increased  ;  then  will  its  perimeter 
coincide  with  the  circumference  of  the  circle,  and 
the  altitude  OG  will  become  equal  to  the  radius,  and 
the  whole  polygon  equal  to  the  circle.  Consequently, 
the  space  of  the  circle,  or  of  the  polygon  in  that  state, 
is  equal  to  the  rectangle  of  the  radius  and  half  the 
circumference.     Q.  E.  D. 

Scholium.  It  will  be  shown  that  the  ratio  of  the  cir- 
cumference of  a  circle  to  its  diameter  may  be  express- 
ed approximately  by  the  mixed  decimal  3.1415926, 
a  number  which  in  all  mathematical  books  it  is  cus- 
tomary to  represent  by  the  Greek  letter  n* 

If  now  d  denote  the  diameter  of  a  circle,  r  the  ra- 
dius, c  the  circumference,  and  a  the  area,  we  shall 
have  the  following  formulae  derived  from  the  forego- 
ing theorems : 

C  =  TTC?     ...  (1) 
C  =  27TT  .  .  .  (2) 

a  =  nr2  ...  (3) 

(1)  is  obtained  by  multiplying  d  by  the  ratio  of  c  arid  d ; 

(2)  "         "         substituting  2r  for  d  in  (1)  ; 

(3)  "  "  multiplying  (2)  by  { r,  in  accordance 
with  th.  73. 

KXERCI8ES. 

1.  Prove  that  no  two  lines  in  a  circle  bisect  each  other  except  two 
diameters. 

2.  Prove  that  lines  drawn  from  the  vertex  of  a  triangle  divide  the 
base  and  a  parallel  to  the  base  proportionally. 

*  For  the  investigation  of  the  ratio  of  the  circumference  of  a  circle 
to  its  diameter,  see  Mensuration,  at  the  last  part  of  this  volume. 


72  GEOMETRY. 

3.  Prove  that  if  a  line  bisect  the  external  angle  of  a  triangle,  the 
distances  of  the  point  in  which  it  meets  the  side  opposite  from  the 
extremities  of  that  side  produced,  will  have  the  same  ratio  as  the 
other  two  sides  of  the  triangle. 

4.  Through  a  given  point,  situated  between  the  sides  of  an  angle, 
to  draw  a  line  terminating  at  the  sides  of  the  angle,  and  in  such  a 
manner  as  to  be  equally  divided  at  the  point. 

5.  Prove  that  the  hypothenuse  of  a  right-angled  triangle  is  to  either 
segment  formed  by  a  perpendicular  upon  the  hypothenuse  from  the 
opposite  vertex,  as  the  square  on  the  hypothenuse  is  to  the  square  on 
the  side  adjacent  the  segment. 

6.  Construct  a  quadrangle  similar  to  a  given  quadrangle,  the  sides 
of  the  latter  having  to  the  former  the  ratio  of  2  to  3. 

7.  Divide  a  line  into  parts  proportional  to  three  given  lines. 

8.  To  draw  a  line  parallel  to  the  base  of  a  given  triangle  in  such  a 
manner  as  to  halve  the  triangle. 

9.  If  from  a  point  in  the  circumference  of  a  circle  two  chords  be 
drawn  to  the  extremities  of  any  diameter  and  a  perpendicular ;  sup- 
posing the  diameter  to  be  20,  and  the  ratio  of  the  segment  into  which 
the  perpendicular  divides  it  2:3,  what  are  the  lengths  of  the 
chords  ? 

10.  To  divide  a  given  line  into  two  parts,  such  that  they  shall  have 
the  ratio  of  two  given  squares. 

11.  To  find  a  line  which  shall  be  to  a  given  line  as  \/£  :  VlO. 

12.  To  construct  a  triangle,  having  given  the  base,  the  vertical  an- 
gle, and  the  ratio  of  the  two  sides  which  contain  it.  (SeePr.21,p.84) 

13.  The  same  with  the  same  ratio,  the  altitude  and  one  of  the  an- 
gles at  the  base  given. 

14.  The  same  with  the  altitude,  the  ratio  of  the  two  segments  of 
the  base,  and  the  vertical  angle. 

15.  The  same  when  the  base,  the  vertical  angle,  and  the  sum  of  the 
squares  of  the  two  sides  are  given. 

16.  The  same  when  the  base,  the  altitude,  and  the  6um  of  the 
squares  of  the  two  sides. 

17.  The  same  with  the  difference  of  the  squares. 

18.  To  inscribe  in  a  given  triangle  another  triangle  of  given  angles 
in  such  a  manner  that  one  of  its  sides  may  be  parallel  to  one  of  the 
sides  of  the  given. 

19.  Also,  when  the  required  triangle  is  required  to  be  similar  to 
the  given,  and  one  of  its  vertices  at  a  given  point  in  one  of  the  sides 
of  the  given  triangle. 

20.  The  same  similar  to  another  instead  of  the  given  triangle. 


EXERCISES. 


73 


21.  To  describe  a  circle  passing  through  a  given  point,  aud  touch- 
ing the  two  aides  of  ■  given  angle. 

22.  Through  two  points  touching  a  given  line. 

23.  Touching  a  line  and  circle,  and  passing  through  a  given  point. 

24.  To  construct  a  square  when  the  difference  between  its  diagonal 
and  side  are  given. 

25.  To  find  a  point  in  a  given  line  from  which,  if  lines  be  drawn 
to  two  given  points  without,  they  shall  have  a  given  ratio. 

20.  Prove  that  if  two  circles  touch  each  other,  the  secants  through 
the  point  of  contact  and  terminating  in  the  two  circumferences  are 
divided  proportionally  at  that  point. 

27.  Prove  that  the  two  common  tangents  and  the  line  joining  the 
centers  of  two  circles  meet  in  the  same  point. 

28.  Draw  a  tangent  to  two  circles  of  different  centers  and  radii. 

29.  Prove  that  if  a  line  be  drawn  through  the  points  of  intersection 
of  two  circles,  tangents  drawn  from  any  point  of  this  line  will  be 
equal. 

30.  That  two  parallelograms  are  similar  when  they  have  an  angle 
in  each  equal  contained  by  proportional  sides. 

31.  Prove  that  the  ratio  of  the  diagonal  to  the  side  of  a  square  is 
that  of  v/2  to  1. 

32.  To  construct  a  polygon  similar  to  a  given  one,  and  bearing  to  it 
a  given  ratio. 

33.  To  construct  a  polygon  similar  to  one  given  polygon  and  equal 
to  another. 

34.  Construct  a  rectangle  equal  to  a  given  square,  having  the  sum 
of  its  sides  equal  to  a  given  line.  The  same,  except  the  difference 
of  the  sides  equal  to  a  given  line. 

35.  To  find  a  point  such  that  the  sum  of  the  squares  of  its  distances 
from  two  given  points  shall  be  equal  to  a  given  square. 

D 


c 

A 

' E    \ 


PROBLEMS. 


PROBLEM   I.* 

To  bisect  a  given  line  AB. 

From  the  two  centers  A  and  B,  with 
any  equal  radii  greater  than  AE,  describe 
arcs  of  circles,  intersecting  each  other 
in  C  and  D  ;  and  draw  the  line  CD,  which 
will  bisect  the  given  line  AB  in  the  point  E. 

For  C  and  D  both  belong  to  the  per- 
pendicular at  the  middle  of  AB  (th.  17,  D 
corol.  1)  ;  and  as  but  one  line  can  be  drawn  through 
two  points,  CD  must  be  this  bisecting  perpendicular. 

PROBLEM    II. 

At  a  given  point  C,  in  a  line  AB,  to  erect  a  perpen- 
dicular. 

From  the  given  point  C,  with  any  ra-              F 
dius,  cut  off  any  equal  parts  CD,  CE  of 
the  given  line ;  and  from  the  two  cen- 
ters D  and  E,  with  any  one  radius,  /___   \__ 

describe  arcs  intersecting  in  F;  then  A  D      c     E  ^ 
join  CF,  which  will  be  perpendicular,  as  required. 

For  the  points  F  and  C  both  belong  to  the  perpen- 
dicular at  the  middle  of  DE,  and  determine  it.  (See 
note  to  Axioms.jf 

*  Many  of  the  following  problems  are  for  the  benefit  of  those  who 
omit  the  exercises. 

t  The  most  convenient  way  of  drawing  perpendiculai-s  through 
points  within  or  without  lines  is  by  means  of  a  rule  and  triangle 
made  of  wood,  metal,  or  any  other  hard  substance. 

The  triangle  is  right-angled,  and  its  hypothenuse  being  placed 
against  the  rule,  with  one  of  its  perpendicular  sides  coinciding  with 
the  given  line,  the  triangle  is  moved  up  or  down  obliquely,  sliding 
along  the  rule,  till  the  other  perpendicular  side  passes  through  the 
given  point ;  a  line  drawn  along  this  latter  side  will  be  the  perpen- 
dicular required. 


PROBLEMS.  75 

OTHERWISE. 

When  the  point  is  near  the  end  of  the  line. 

Analysis.*  Suppose  the  perpendic-  F 

ular  CF  drawn ;  FCA  will  then  be  a 
right  angle.  But  we  know  that  a  right 
angle  is  inscribed  in  a  semicircle. 
Hence  the  following  construction.  — 

From  any  point  D,  assumed  above  A  E  c  B 
the  line,  as  a  center,  through  the  given  point  C  de- 
scribe a  circle,  cutting  the  given  line  at  E ;  and 
through  E  and  the  center  D  draw  the  diameter  EDF  ; 
then  join  CF,  which  will  be  the  perpendicular  re- 
quired. 

Synthesis.  For  the  angle  at  C,  being  an  angle  in  a 
semicircle,  is  a  right  angle,  and  therefore  the  line  CF 
is  a  perpendicular  (by  def.  12). 


problem  in. 

From  a  given  point  A,  to  let  fall  a  perpendicular  on 
a  given  line  BC. 

From  the  given  point  A  as  the  center,  a 

with  any  convenient  radius,  describe  an  /\ 

arc,  cutting  the  given  line  at  the  two 
points  D  and  E  ;  and  from  the  two  cen-    _>%! 
ters  D,  E,  with  any  radius,  describe  two    B  VT  /  w 
arcs,  intersecting  at  F ;  then  draw  AGF,  \|  / 

which  will  be  perpendicular  to  BC,  as  f 

required. 

With  the  rule  and  triangle,  parallel  lines  to  a  given  line  may  be 
drawn  through  given  points  in  an  obvious  manner. 

Another  method  is  by  means  of  what  is  called  a  T  rule,  from  its 
resemblance  to  this  letter,  the  cross-piece  being  thicker  than  the 
other,  so  as  to  project  below  the  edge  of  a  rectangular  drawing  board, 
upon  which  the  paper  is  pasted.  The  lines  drawn  with  this  will  bo 
always  parallel  to  the  edges  of  the  board,  unless  there  be  a  move- 
ment of  one  arm  of  the  rule,  so  that  it  may  be  placed  at  any  angle 
with  the  other. 

*  The  student  may  be  exercised  in  giving  the  analysis  of  some 
others  of  the  problems. 


76 


GEOMETRY. 


For  the  points  A  and  F  are  both  equally  distant 
from  the  points  D  and  E ;  hence  AF  is  perpendicu- 
lar at  the  middle  of  DE. 


OTHERWISE. 

When  the  point  is  nearly  the  opposite  end  of  the  line. 

From  any  point  D,  in  the  given  A 

line  BC,  as  a  center,  describe  the  arc 
of  a  circle  through  the  given  point 
A,  cutting  BC  in  E  ;  and  from  the  b- 
center  E,  with  the  radius  EA,  de- 
scribe another  arc,  cutting  the  for- 
mer in  F;  then  draw  AGF,  which 
will  be  perpendicular  to  BC,  as  required. 

For  the  chords  AE,  EF  being  equal,  their  arcs 
are  equal,  and  the  line  DE,  drawn  through  the  center 
D  and  middle  point  E  of  the  arc  AEF,  is  perpendicu- 
lar to  the  chord  AF  of  that  arc.     (See  th.  34.) 


PROBLEM    IV. 

To  bisect  a  given  angle. 

Let  ACB  be  the  given  angle. 
With  C  as  a  center,  describe  an 
arc,  cutting  the  sides  of  the  given 
angle  in  A  and  B.  Draw  the  chord 
AB,  and  from  C  the  perpendicular 
CD  to  this  chord,  which  (th.  34)  will 
bisect  the  given  angle. 


PROBLEM   V. 

To  make  a  triangle  with  three  given  lines  AB,  AC,  BC. 

With  the  center  A,  and  distance  AC,  de- 
scribe an  arc.  With  the  center  B,  and  dis- 
tance BC,  describe  another  arc,  cutting  the 
former  in  C.  Draw  AC,  BC,  and  ABC  will 
be  the  triangle  required. 


For  the  radii,  or  sides  of  the  triangle,  b c 


PROBLEMS.  77 

AC,  BC  are  equal  to  the  given  lines  AC,  BC  by  con- 
struction. 

Note.  If  any  two  of  the  lines  are  not  together 
greater  than  the  third,  the  construction  is  impossible. 

PROBLEM   VI. 

At  a  given  point  A,  in  a  line  AB,  to  make  an  angle 
equal  to  a  given  angle  C. 

From  the  centers  A  and  C,  with  any 
one  radius,  describe  the  arcs  DE,  BF. 
Then,  with  radius  DE,  and  center  B, 
describe  an  arc,  cutting  BF  in  G. 
Through  G  draw  the  line  AG,  and  it  c 
will  form  the  angle  required. 

Let  the  equal  lines  or  radii,  DE,  Syk 

BG,  be  drawn.     Then  the  two  trian- 
gles CDE,  ABG,  being  mutually  equi- 
lateral, are  mutually  equiangular  (th.    ■* 
5),  and  have  the  angle  at  A  equal  to  the  angle  at  C* 

PROBLEM   VII. 

Through  a  given  point  A,  to  draw  a  line  parallel  to 
a  given  line  BC. 

From  the  given  point  A  draw   B  D      r 

a  line  AD  to  any  point  in  the    ~p ' 

given  line  BC.     Then  draw  the  / 

line  AE,  making  the  angle  at  A    -£- — 

equal  to  the  angle  at  D  (by  prob. 

6) ;  so  shall  AE  be  parallel  to  BC,  as  required. 

*  Angles  are  made  most  conveniently  with  a  protracter,  which  is 
commonly  a  semicircle  of  metal,  horn,  or  paper,  divided  into  degreM 
and  parts  of  a  degree.  An  angle  equal  to  a  given  angle  is  protracted 
by  placing  the  diameter  of  the  instrument  upon  one  side  of  the  given 
angle,  the  center  being  at  the  vertex,  and  then  the  other  side  of  the 
given  angle  will  pass  through  the  number  of  the  degrees  which  it 
contains.  If,  then,  the  protracter  be  taken  up  and  placed  with  its 
diameter  upon  the  given  line,  and  center  at  the  given  point,  and  a 
point  marked  on  the  paper  at  the  same  degree  or  division  of  the  cir- 
cumference, and  this  point  joined  with  the  given  point,  the  required 
angle  will  be  formed. 


78  GEOMETRY. 

For,  the  angle  D  being  equal  to  the  alternate  angle 
A,  the  lines  BC,  AE  are  parallel,  by  th.  10. 

FROBLEM    VIH. 

Given  two  sides  of  a  triangle  and  the  angle  opposite 
one  of  them  to  construct  the  triangle. 

There  will  be  two  cases  : 

1.  Where  the  given  triangle  is 
right  or  obtuse.  Draw  two  lines, 
AB,  AC,  making  with  each  other 
the  given  angle.  Take  AC,  equal  A 
to  one  of  the  given  sides,  and  with  C  as  a  center  and 
the  given  side  opposite  the  given  angle  as  a  radius, 
cut  the  indefinite  side  AB  in  B.  Join  CB,  and  ABC 
will  be  the  triangle  constructed  with  the  given  angle 
and  sides. 

2.  If  the  given  angle  be 
acute,  and  the  side  opposite 
be  less  than  the  other  given 
side,  there  will  be  two  solu- 
tions. The  triangle  ABF  or 
ABE,  either  of  them  being  B 
constructed  with  the  given  angle  B  and  side  AE  or 
AF.    This  is  called  a  doubtful  or  ambiguous  solution. 

If  AE  or  AF  be  just  long  enough  to  reach  BC, 
the  two  solutions  coalesce,  and  the  resulting  triangle 
ADB  is  right-angled  at  D.  If  AF  is  too  short  to  reach 
BC,  the  solution  is  impossible. 

N.B. — With  the  exception  of  this  case,  it  may  be 
said,  as  a  general  scholium,  that  a  triangle  is  determ- 
ined when  any  three  parts*  are  given,  one  of  which 
is  a  side.  Or,  two  triangles  are  equal  when  any 
three  parts,  one  of  which  is  a  side  in  the  one,  are 
equal  to  the  same  in  the  other. 

*  The  parts  of  a  triangle  are  the  three  sides  and  the  three  angles — 
six  in  all. 


/ 


PROBLEMS. 


79 


PROBLEM    IX. 

To  divide  a  line  AB  into  any  proposed  number  of 
equal  parts. 

Draw  any  other  line  AC, 
forming  any  angle  with  the 
given  line  AB  ;  on  which  set 
off  any  line  AD  as  many  times 
as  there  are  to  be  parts  in  AB 
ending  at  C.  Join  BC,  parallel 
to  which  draw  DE,  then  AE  A 
will  apply  exactly  the  required  number  of  times  to 
AB.  For  those  parallel  lines  divide  both  the  sides 
AB,  AC  proportionally,  by  th.  61. 


problem  x. 
To  make  a  square  on  a  given  line  AB. 

Raise  AD,  BC,  each  perpendicular  and  d 
equal  to  AB,  and  join  DC  ;  so  shall  ABCD 
be  the  square  sought. 

For  all  the  three  sides  AB,  AD,  BC  are 
equal,  by  the  construction,  and  DC  is  equal 
and  parallel  to  AB  (by  th.  21)  ;  so  that  all 
the  four  sides  are  equal,  and  the  opposite  ones  are 
parallel.  Again,  the  angle  A  or-  B  of  the  parallelo- 
gram, being  a  right  angle,  the  angles  are  all  right 
ones  (cor.  1,  th.  19).  Hence,  then,  the  figure,  having 
all  its  sides  equal  and  all  its  angles  right,  is  a  square 
(def.  32). 

PROBLEM    XI. 

To  make  a  rectangle  or  a  parallelogram  of  a  given 
length  and  breadth,  AB,  BC. 

Erect  AD,  BC  perpendicular  to  AB,  and  D  c 

each  equal  to  BC ;  then  join  DC,  and  it  is 
done. 

The  demonstration  is  the  same  as  in  the  A 
last  problem.  B 0 


80  GEOMETRY. 

And  in  the  same  manner  is  described  any  oblique 
parallelogram,  only  drawing  AD  and  BC  to  make  the 
given  oblique  angle  with  AB,  instead  of  perpendicu- 
lar to  it. 

PROBLEM    XII. 

To  make  a  rectangle  equal  to  a  given  triangle  ABC. 
Bisect  the  base  AB  in  D ;  then  raise       c  E       f 
DE  and  BF  perpendicular  to  AB,  and 
meeting  CF  parallel  to  AB  at  E  and  F ; 
so  shall  DF  be  the  rectangle  equal  to  the 
given  triangle  ABC  (by  cor.  of  th.  23).      a       D        u 

PROBLEM    XIII. 

To  make  a  square  equal  to  the  sum  of  two  or  more 
given  squares. 

Let  AB  and  AC  be  the  sides  of  two 
given  squares.  Draw  two  indefinite 
lines,  AP,  AQ,  at  right  angles  to  each 
other,  in  which  place  the  sides  AB, 
AC  of  the  given  squares ;  join  BC : 
then  a  square  described  on  BC  will  be 
equal  to  the  sum  of  the  two  squares 
described  on  AB  and  AC  (th.  26). 

In  the  same  manner,  a  square  may  be  made  equal 
to  the  sum  of  three  or  more  given  squares.  For,  if 
AB,  AC,  AD  be  taken  as  the  sides  of  the  given 
squares,  then,  making  AE  =  BC,  AD  =  AD,  and 
drawing  DE,  it  is  evident  that  the  square  on  DE  will 
be  equal  to  the  sum  of  the  three  squares  on  AB,  AC, 
AD.     And  so  on  for  more  squares. 

PROBLEM   xiv. 

To  make  a  square  equal  to  the  difference  of  two  given 
squares. 

Let  AB   and  AC,  taken  in  the   same        - — ^jp 
straight  line,  be  equal  to  the  sides  of  the    f        /\ 

two  given  squares.     From  the  center  A,    ' /.  \\ 

with  the  distance  AB,  describe  a  circle, 
and  make  CD  perpendicular  to  AB,  meeting  the  cir- 
cumference in  D :  so  shall  a  square  described  on  CD 


PROBLEMS. 


81 


be  equal  to  ADa 
(cor.  1,  th.  26). 


AC3,  or  ABa  —  AC2,  as  required 


PROBLEM    XV. 

To  make  a  triangle  equal  to  a  given  polygon  ABCDE. 

Draw  DB  and  CF  parallel  to  it, 
meeting  AB  produced  at  F  ;  then 
draw  DF ;  so  shall  the  polygon 
DFAE  be  equal  to  the  given  pol- 
ygon ABCDE. 

For  the  triangle  DFB  -  DCB 
(th.  22) ;  therefore,  by  adding 
DBAE  to  the  equals,  the  sums  are  equal  (ax.  2),  that 
is,  DBAE  +  DBF  =  DBAE  +  DCB,  or  the  quad- 
rilateral DFAE  =  to  the  pentagon  ABCDE. 

In  a  similar  manner  the  number  of  sides  of  a  pol- 
ygon may  be  repeatedly  reduced,  by  one  each  time, 
till  the  polygon  is  changed  into  an  equivalent  triangle. 


..-•'" 


PROBLEM    XVI. 

To  make  a  square  equal  to  a  given  rectangle 
Let  AB,  BC  be  equal  to  the  ad-  ^ 

jacent  sides  of  the  rectangle. 

Produce  one  side  AB  till  BC  be 
equal  to  the  other  side.  On  AC 
as  a  diameter  describe  a  circle 
meeting  the  perpendicular  BD  at  A  B       C 

D ;  then  will  BD  be  the  side  of  the  square  equal  to 
the  given  rectangle,  as  appears  by  cor.  1,  th.  66. 


N 


PROBLEM   XVII. 

To  describe  a  circle  about  a  given  triangle  ABC. 

Bisect  any  two  sides  with  two  of  the 
perpendiculars  FD,  ED,  or  GD,  and 
the  point  D,  in  which  they  intersect, 
will  be  the  center. 

For  every  point  of  the  line  FD  must 
be  equally  distant  from  the  points  B 
and  C  (th.  17,  corol.  1),  and  every  point 
D2 


82  GEOMETRY. 

of  the  line  ED  must  be  equally  distant  from  A  and  B  ; 
hence  the  point  D,  common  to  these  two  lines,  must 
be  at  equal  distances  from  the  three  points  A,  B,  and 
C,  and  the  center  of  a  circle  passing  through  them. 

Scholium.  There  is  but  one  such  circle.  For  its 
center  could  not  be  out  of  the  line  FA  (th.  18),  nor  out 
of  EC,  and  they  intersect  in  but  one  point  D.* 

Note.  The  problem  is  the  same,  in  effect,  when  it  is 
required 

To  describe  the  circumference  of  a  circle  through 
three  given  points  A,  B,  0,  or  to  find  the  center  of  a 
given  circle  or  arc. 

Draw  chords  BA,  BC,  and  bisect  these  chords  per- 
pendicularly by  lines  meeting  in  D,  which  will  be 
the  center.    (See  last  diagram.) 

PROBLEM    XVIII. 

An  isosceles  triangle  ABC  being  given,  to  describe 
another  on  the  same  base  AB,  whose  vertical  angle  shall 
be  only  half  the  vertical  angle  C. 

From  C  as  a  center,  with  the  dis- 
tance CA,  describe  the  circle  ABE. 
Bisect  AB  in  D,  join  DC,  and  pro- 
duce to  the  circumference  at  E;  join 
EA  and  EB,  and  ABE  shall  be  the 
isosceles  triangle  required. 

For  every  point  of  the  perpendic- 
ular DE  is  equally  distant  from  A  and  B  (th.  17,  co- 
rol.  1)  ;  hence  the  side  EA  must  be  equal  to  the  side 
EB  of  the  triangle  AEB,  which  is,  therefore,  isosceles, 
and  the  angle  ACB  at  the  center  must  be  double  of 
the  angle  AEB  at  the  circumference,  for  they  both 
stand  on  the  same  segment  AB. 

*  If  the  given  triangle  be  acute  angled,  the  center  of  the  circle 
will  be  within  it;  and  if  the  triangle  be  equilateral,  as  in  the  dia- 
gram, the  center  of  the  circle  will  be  the  center  of  the  triangle,  and 
the  perpendiculars  at  the  middle  of  the  sides  will  pass  through  the 
vertices  of  the  opposite  angles. 

If  the  triangle  be  obtuse  angled,  the  center  of  the  circle  will  fall 
without;  if  right  angled,  the  center  will  fall  upon  the  hypothenuse. 


PROBLEMS. 


83 


PROBLEM   XIX. 

Given  an  isosceles  triangle  AEB,  to  erect  another  on 
the  same  base  AB,  which  shall  have  double  the  vertical 
angle  E. 

Describe  a  circle  about  the  triangle 
AEB,  find  its  center  C,  and  join  CA, 
CB,  and  ACB  is  the  triangle  required. 

The  angle  C  at  the  center  is  double 
of  the  angle  E  at  the  circumference, 
and  the  triangle  ACB  is  isosceles  ;  for 
the  sides  CA,  CB,  being  radii  of  the 
same  circle,  are  equal. 


PROBLEM   xx. 

To  draw  a  tangent  to  a  circle,  through  a  given  point  A. 

1.  When  the  given  point  A  is  in  the 
circumference  of  the  circle,  join  A 
and  the  center  O  ;  perpendicular  to 
which  draw  BAC,  and  it  will  be  the 
tangent,  by  th.  36. 

2.  When  the  given  point  A  is  out 
of  the  circle,  draw  AO  to  the  center 
O  ;  on  which,  as  a  diameter,  describe 
a  semicircle,  cutting  the  given  cir- 
cumference in  D  ;  through  which 
draw  BADC,  which  will  be  the  tangent,  as  required. 

For  join  DO.  Then  the  angle  ADO,  in  a  semicir- 
cle, is  a  right  angle,  and,  consequently,  AD  is  perpen- 
dicular to  the  radius  DO,  or  is  a  tangent  to  the  circle 
(th.  36). 

Scholium.  The  circle  ADO  cuts  the  given  circle  in 
two  points ;  and  there  will  be  two  tangents,  AD 
and  AE,  to  the  given  circle  from  the  same  point  A, 
without.  These  tangents  are  equal  in  length,  and  the 
line  joining  the  point  without  and  the  center  bisects 
the  angle  which  the  tangents  make  with  each  other ; 
for  the  right-angled  triangles  ADO,  AEO,  having  the 


84 


GEOMETltV. 


side  OA  common,  and  the  side  OD  =  OE  being  radii 
of  the  same  circle,  the  triangles  are  equal  .*.  AD  = 
AE  and  angle  DAO  =  angle  EAO. 

PROBLEM    XXI. 

On  a  given  line  AB  to  describe  a  segment  of  a  circle 
capable  of  containing  a  given  angle. 

At  the  ends  of  the  given  line  make 
angles  DAB,  DBA,  each  equal  to  the 
given  angle  C.  Then  draw  AE,  BE 
perpendicular  to  AD,  BD  ;  and  with  theA/ 
center  E,  and  radius  EA  or  EB,  describe 
a  circle ;  so  shall  AFB  be  the  segment 
required,  as  any  angle  F  made  in  it  will 
be  equal  to  the  given  angle  C. 

For  the  two  lines  AD,  BD,  being  perpendicular  to  the 
radii  EA,  EB  (by  construction),  are  tangents  to  the 
circle  (th.  36)  ;  and  the  angle  A  or  B,  which  is  equal 
to  the  given  angle  C  by  construction,  is  equal  to  the 
angle  F,  being  all  three  measured  by  half  the  arc  AB 
(th.  38  and  39).* 

Scholium.  One  of  the  lines  AD,  BD  may  be  omit- 
ted, and  a  perpendicular  drawn  at  the  middle  of  AB 
to  meet  the  other  at  the  point  E. 

*  This  problem  is  particularly  useful  in  the  survey  of  harbors. 
Three  points  on  the  shore  are  chosen,  which,  being  connected  by 
lines,  form  a  triangle  ;  then  from  a  boat,  where  a  sounding  is  to  be 
made,  the  angles  subtended  by  two  of  the  sides  of  this  triaugle  are 
measured  with  a  sextant. 

To  transfer  this  to  a  map,  there  must  first  be  made  upon  the  paper 
the  triangle  whose  sides  unite  the  three  points  upon  the  shore.  Then 
upon  one  of  the  sides  of  this  triangle,  by  the  above  problem,  make  a 
segment  capable  of  containing  one  of  the  observed  angles,  and  upon 
the  other  a  segment  capable  of  containing  the  other  observed  angle  ; 
the  point  in  which  the  arcs  of  these  two  segments  intersect  will  be  the 
point  on  the  map  corresponding  to  that  where  the  sounding  was 
made,  and  there  the  depth  in  fathoms  or  feet  may  be  written  down. 


TR0BLEMS. 


85 


PROBLEM    XXII. 

To  inscribe  an  equilateral  triangle  in  a  given  circle. 

Through  the  center  C  draw  any  di-  A 

ameter  AB.  From  the  point  B  as  a 
center,  with  the  radius  BC  of  the  given 
circle,  describe  an  arc  DCE.  Join  AD, 
AE,  DE,  and  ADE  is  the  equilateral 
triangle  sought. 

Join  DB,  DC,  EB,  EC.  Then  DCB 
is  an  equilateral  triangle,  having  each  side  equal  to 
the  radius  of  the  given  circle.  In  like  manner,  BCE 
is  an  equilateral  triangle.  But  the  angle  ADE  is 
equal  to  the  angle  ABE  or  CBE,  standing  on  the  same 
arc  AE ;  also,  the  angle  AED  is  equal  to  the  angle 
CBD,  on  the  same  arc  AD ;  hence  the  triangle  DAE 
has  two  of  its  angles,  ADE,  AED,  equal  to  the  an- 
gles of  an  equilateral  triangle,  and  therefore  the  third 
angle  at  A  is  also  equal  to  the  same ;  so  that  the  tri- 
angle is  equiangular,  and  therefore  equilateral. 


PROBLEM    XXIII. 

To  inscribe  a  circle  in  a  given  triangle  ABC. 

Bisect  any  two  angles  C 
and  B  with  the  two  lines  CD, 
BD.  From  the  intersection  D, 
w(iich  will  be  the  center  of  the 
circle,  draw  the  perpendiculars 
DE,  DF,  DG,  and  they  will  be 
the  radii  of  the  circle  required. 

For,  since  the  sides  CB,  CA, 
are  to  be  tangents,  the  line  CD, 
bisecting  the  angle  which  they  form,  must  pass  through 
the  center.  (Prob.  20,  schol.)  For  a  similar  reason, 
BD  must  pass  through  the  center.  Hence  it  is  at  the 
intersection  D  of  these  two  lines. 


86  GEOMETRY. 


PROBLEM    XXIV. 

To  inscribe  a  square  in  a  given  circle. 

Draw  two  diameters  AC,  BD,  je 

crossing  at  right  angles  in  the  cen-  S* 

ter  E.     Then  join  the  four  extrem-       /  / 
ities   A,  B,  C,  D  with  right  lines,    A/ 
and  these  will  form  the  inscribed      \\ 
square  ABCD.  V\ 

For  the  four  right-angled  trian-  \ 

gles  AEB,  BEC,  CED,  DEA  are  B 

identical,  because  they  have  the  sides  EA,  EB,  EC, 
ED  all  equal,  being  radii  of  the  circle,  and  the  four 
included  angles  at  E  all  equal,  being  right  angles,  by 
the  construction.  Therefore,  all  their  third  sides,  AB, 
BC,  CD,  DA,  are  equal  to  one  another,  and  the  figure 
ABCD  is  equilateral.  Also,  all  its  four  angles,  A,  B, 
C,  D,  are  right  ones,  being  angles  in  a  semicircle. 
Consequently,  the  figure  is  a  square. 

PROBLEM   XXV. 

To  find  a  fourth  proportional  to  three  given  linesf 

AB,  AC,  AD. 

Place  two  of  the  given  lines  AB,  A_^ B 

AC,  or  their  equals,  to  make  any  an-  a -c 

gle  at  A  ;  and  on  AB  set  off,  or  place,  A d 

the  other  line  AD,  or  its  equal.     Join  J<^\ 

BC,  and  parallel  to  it  draw  DE ;  so       -- \   \ 

shall  AE  be  the  fourth  proportional,  as 

required. 

For,  because  of  the  parallels  BC,  DE,  the  two 
sides  AB,  AC  are  cut  proportionally  (th.  61)  ;  so  that 
AB  :  AC  : :  AD  :  AE. 

PROBLEM   XXVI. 

T<  find  a  mean  proportional  between  two  lines  AB, 
BC. 


PROBLEMS.  87 

Place  AB,  BC,  joined  in  one  straight  A B 

line  AC ;  on  which,  as  a  diameter,  de-  B c 

scribe   the   semicircle  ADC  ;  to  meet         ^ — £ 
which  erect  the  perpendicular  BD,  and 
it  will  be  the  mean  proportional  sought     i 
between  AB  and  BC  (by  cor.  1,  th.  66).     * 


O  B     C 


PROBLEM    XXVII. 

To  make  a  square  equal  to  a  given  triangle. 

Find  a  mean  proportional  between  the  base  and 
half  the  altitude,  or  between  the  altitude  and  half  the 
base  of  the  triangle,  and  it  will  be  the  side  of  the 
square  required. 

Corol.  To  find  a  square  equal  to  a  given  polygon, 
first  find  a  triangle  equal  to  the  given  polygon  by 
Prob.  15,  and  then  a  square  equal  to  the  given  trian- 
gle.    This  is  called  quadrating  the  polygon.* 

PROBLEM    XXVIII. 

To  divide  a  given  line  in  extreme  and  mean  ratio. 

Let  AB  be  the  given  line 
to  be  divided  in  extreme  and 
mean  ratio,  that  is,  so  that  the  JJ* 

whole  line  may  be  to  the  great-  J), 

er  part  as  the  greater  is  to  the 
less  part. 

Draw  BC  perpendicular  to  A  E        B 

AB,  and  equal  to  half  AB.  Join  AC  ;  and  with  cen- 
ter C  and  distance  CB,  describe  the  circle  BD  ;  then 
with  center  A  and  distance  AD,  describe  the  arc 
DF  ;  so  shall  AB  be  divided  in  F  in  extreme  and 
mean  ratio,  or  so  that  AB  :  AF  :  :  AF  :  FB. 

For  produce  AC  to  the  circumference  at  E.  Then, 
ADE  being  a  secant,  and  AB  a  tangent,  because  B  is 
a  right  angle ;  therefore  the  rectangle  AE'AD  is  equal 
to  AB2  or  AB-AB  (cor.  1,  th.  42) ;  taking  the  first  as 

*  The  quadrature  of  the  circle  has  occupied  ingenious  minds  in  a 
fi  uiilr>s  undertaking  from  a  remote  antiquity.  The  impossibility  of 
this  problem  may  be  very  satisfactorily  proved. 


88  GEOMETRY. 

means,  and  second  as  extremes  of  a  proportion  (th. 
54),  we  have  AB  :  AE  or  AD  4-  DE  :  :  AD  :  AB. 
Bat  AF  is  equal  to  AD,  by  construction,  and  AB  =  2 
BC  =  DE ;  therefore,  AB  :  AF  +  AB  :  :  AF  :  AB 
or  AF  +  FB ;  and,  by  division  (th.  47),  AB  :  AF  :  : 
AF  :  FB. 

PROBLEM    XXIX. 

To  describe  a  regular  pentagon  on  a  given  line  AB. 

On  AB  erect  the  isosceles  trian-  q 

gle  ACB,  having  each  of  the  an- 
gles at  the  base  double  of  its  verti- 
cal angle  ;#  on  AB  again  construct  D< 
another  isosceles  triangle  whose 
vertical  angle  AOB  is  double  of  A 
CB,  and  about  the  vertex  O  place 
the  isosceles  triangles  AOD,  DOC, 
COE,*and  EOB,  each  =  AOB  ;  these  triangles  will 
compose  a  regular  pentagon. 

For  the  angle  AOB,  being  the  double  of  ACB, 
which  is  the  fifth  part  of  two  right  angles,  must  be 
equal  to  the  fifth  part  of  four  right  angles ;  and,  con- 
sequently, five  angles,  each  of  them  equal  to  AOB, 
will  adapt  themselves  about  the  point  O.  But  the 
bases  of  those  central  triangles,  and  which  form  the 
sides  of  the  pentagon,  are  all  equal ;  and  the  angles 
at  their  bases  being  likewise  equal,  they  are  equal  in 
the  collective  pairs  which  constitute  the  internal  an- 
gles of  the  figure.  It  is,  therefore,  a  regular  pen- 
tagon. 

PROBLEM   XXX. 

To  describe  a  hexagon  upon  a  given  line  AB. 

From  A  and  B  as  centers,  with  AB  as  radius,  de- 
scribe arcs  intersecting  in  O  (fig.  to  the  next  problem). 
From  O  as  a  center,  with  the  same  radius,  describe  a 
circle  ABCDEF.     Within  this  circle  set  off  from  B 

*  In  the  last  diagram,  AB  being  the  given  base.  AE  will  be  the 
side  of  the  isosceles  triangle.     (See  Prob.  33.) 


PROBLEMS. 


81) 


the  chords  BC,  CD,  DE,  EF,  FA  in  succession,  each 
equal  to  AB  :  they  will,  together  with  AB,  form  the 
bexagon  required. 

The  demonstration  is  analogous  to  that  of  the  fol- 
lowing problem. 

PROBLEM    XXXI. 

To  inscribe  a  regular  hexagon  in  a  circle. 

Apply  the  radius  AO  of  the  given 
circle  as  a  chord,  AB,  BC,  CD,  &c, 
quite  round  the  circumference,  and  it 
will  complete  the  regular  hexagon  AB 
CDEF. 

For,  draw  the  radii  AO,  BO,  CO,  DO, 
EO,  FO,  completing  six  equal  triangles  ; 
of  which  any  one,  as  ABO,  being  equilateral  (by  con- 
str.),  its  three  angles  are  all  equal  (cor.  2,  th.  3),  and 
any  one  of  them,  as  AOB,  is  one  third  of  the  whole, 
or  of  two  right  angles  (th.  15),  or  one  sixth  of  four 
right  angles.  But  the  whole  circumference  is  the 
measure  of  four  right  angles  (cor.  4,  th.  6).  There- 
fore the  arc  AB  is  one  sixth  of  the  circumference  of 
the  circle,  and,  consequently,  its  chord  AB  one  side 
of  an  equilateral  hexagon  inscribed  in  the  circle. 

Cor.  1.  The  chord  of  00°  is  equal  to  the  radius  of 
the  circumscribing  circle. 

Cor.  2.  To  inscribe  an  equilateral  triangle,  join  the 
alternate  vertices  of  the  inscribed  hexagon. 

PROBLEM    XXXII. 

On  a  given  line  AB  to  construct  a  regular  octagon. 
Bisect  AB  by  the  perpendic- 
ular CD,  which  make  =  CA  or 
CB  ;  join  DA  and  DB  ;  produce 
CD,  making  DO  =  DA  or  DB, 
draw  AO  and  BO,  thus  forming 
an  angle  equal  to  the  half  of  AD 
B  (Prob.  18),  and  about  the  ver- 
tex O  repeat  the  equal  triangles 
AOB,  AOE,  EOF,  FOG,  GOH,  a    o    b 

HOI,  IOK,  and  KOB  to  compose  the  octagon. 


90  GEOMETRY. 

For  CA,  CD,  and  CB  being  all  equal  by  construc- 
tion, the  angle  ADB  is  contained  in  a  semicircle,  and 
is,  therefore,  a  right  angle.  Consequently,  AOB  is 
equal  to  the  half  of  a  right  angle,  and  eight  such  an- 
gles will  adapt  themselves  about  the  point  O.  Whence 
the  figure  BAEFGHIK,  having  eight  equal  sides  and 
equal  angles,  each  angle,  as  ABK,  being  the  double 
of  ABO,  is  a  regular  octagon. 

PROBLEM    XXXIII. 

To  inscribe  a  regular  decagon  in  a  circle. 

Divide  the  radius  into  extreme 
and  mean  ratio,  and  the  greater 
segment  will  apply  to  the  circum- 
ference ten  times.  For  let  D  be 
the  point  of  division  on  the  ra- 
dius. Then,  since  we  have,  by 
construction, 

CA  :  AB  : :  AB  :  AD, 
the    two   triangles   CAB,   DAB 
have  the  angle  A  common,  and  A         B 

the  sides  about  the  common  angle  proportional ;  they 
are,  therefore  (th.  65),  similar.  But  CAB  is  isosceles, 
CA  and  CB  being  radii ;  therefore,  ADB  is  isosceles 
or  AB  =  BD ;  and  since,  by  construction,  AB  =  CD 
.*.  BD  =  DC,  and  the  triangle  DCB  is  isosceles,  and 
hence  the  angle  C  =  CBD.  But  the  exterior  angle 
ADB  of  the  triangle  DCB  is  equal  to  the  sum  of  the 
two  interior  and  opposite  (th.  13)  ;  or,  since  these  are 
equal,  the  angle  ADB,  or  its  equal  DAB  =  CBA,  is 
equal  to  double  the  angle  C.  Hence  the  triangle  CAB 
is  such  that  the  angles  A  and  B  at  the  base  are  each 
double  the  angle  C  at  the  vertex,  or  together  are  four 
times  the  angle  C ;  or  all  three  of  the  angles  of  the 
triangle  CAB  are  together  equal  to  five  times  the  an- 
gle C.  Hence  the  angle  C  =  one  fifth  of  two  right 
angles  =  one  tenth  of  four  right  angles,  and  the  arc 
AB,  therefore,  which  measures  the  angle  C,  is  one 
tenth  the  circumference. 


PROBLEMS.  91 

CoroL  1.  By  joining  the  alternate  vertices  of  the 
decagon,  a  pentagon  may  be  inscribed. 

CoroL  2.  A  pentedecagon  may  be  inscribed  by  first 
finding  the  arc  of  a  decagon,  then  the  arc  of  a  hexa- 
gon, and  the  difference  between  them  will  be  the  arc 
of  the  figure  required. 

For  i— TV  =  t't- 

PROBLEM    XXXIV. 

To  describe  a  circle  about  a  regular  polygon. 

Bisect  any  two  of  the  angles  C  and 
D  with  the 'lines  CO,  DO;  then  their 
intersection  O  will  be  the  center  of  the 
circumscribing  circle  ;  and  OC  or  OD 
will  be  the  radius. 

For,  draw  OB,  OA,  OE,  &c.f  to  the  ^B 

angular  points  of  the  given  polygon.  Then  the  tri- 
angle OCD  is  isosceles,  having  the  angles  at  C  and  D 
equal,  being  the  halves  of  the  equal  angles  of  the 
polygon  BCD,  CDE ;  therefore,  their  opposite  sides 
CO,  DO  are  equal  (th.  4).  But  the  two  triangles 
OCD,  OCB,  having  the  two  sides  OC,  CD  equal  to 
the  two  OC,  CB,  and  the  included  angles  OCD,  OCB 
also  equal,  will  be  identical  (th.  1),  and  have  their 
third  sides  BO,  OD  equal. 

AO  may  be  proved  equal  to  BO  in  the  same  man- 
ner that  BO  was  proved  equal  to  CO,  and  so  on  ;  and 
thus  the  point  O  be  shown  to  be  equidistant  from  all 
the  vertices  of  the  polygon. 

PROBLEM   XXXV. 

To  inscribe  a  circle  in  a  regular  polygon. 

Bisect  any  two  sides  of  the  polygon 
by  the  perpendiculars  GO,  FO,  and 
their  intersection  O  will  be  the  center 
of  the  inscribed  circle,  and  OG  or  OF 
will  be  the  radius. 

For  if  a  circle  be  circumscribed 
about  the  polygon,  the  perpendiculars 


92  GEOMETRY. 

GO,  FO,  at  the  middle  of  the  chords,  will  meet  in  its 
center  O  (schol.  to  th.  34),  and  the  distances  OG,  OF, 
&c,  of  these  chords  are  equal  (th.  35). 

PROBLEM   XXXVI. 

On  a  given  line  to  construct  a  rectilinear  figure  sim- 
ilar to  a  given  rectilinear  figure. 

Let   abcde  be  the  given  d 

rectilinear  figure,  and  AB  /f\^  d 

the  side  of  the  proposed  / j  ^}c  /P^*% 
similar  figure  that  is  simi-  /  I  /'  t/ j  /j 
larly  posited  with  ab.  \  //    /         \Lr     j 

Place  AB  in  the  prolong-      V  I  ^ Jb 

ation  of  ab,  or  parallel  to  it.  A  B 
Draw  AC,  AD,  AE,  &c.  parallel  to  ac,  ad,  ae,  respect- 
ively. Draw  BC  parallel  to  be,  meeting  AC  in  C  ; 
CD  parallel  to  cd,  and  meeting  AD  in  D  ;  DE  parallel 
to  de,  and  meeting  AE  in  E ;  and  so  on  till  the  figure 
is  completed.  Then  ABCDE  will  be  similar  to  abcde, 
from  the  nature  of  parallel  lines  and  similar  figures 
(th.  68). 

Otherwise,  divide  the  given  figure  up  into  triangles 
as  in  the  diagram  ;  then  upon  ab  make  the  triangle 
abc  equiangular  with  the  triangle  ABC,  and  upon  ac, 
acd,  equiangular  with  ACD,  and  so  on  till  the  figure 
is  completed.     (See  the  demonstr.  of  th.  68.) 


PROBLEMS.  93 


GENERAL  NOTE  UPON  THE   METHOD  OF   SOLUTION  OF 
PROBLEMS. 

Lv  every  problem  of  Plane  Geometry,  it  is  necessary  to  trace  upon  a 
plane,  in  accordance  with  given  conditions,  one  or  more  right  lines 
or  curves,  one  or  more  angles,  one  or  more  points. 

The  problem  can  be  solved  by  the  aid  of  the  rule  and  compass,  if 
the  entire  system  of  lines  to  be  traced,  and  the  lines  of  construction, 
are  reduced  to  a  system  of  right  lines  and  circumferences  of  circles. 

A  line  is  determined  when  two  of  its  points  are  known  ;  a  circum- 
ference when  its  center  and  a  point  of  it,  or  when  three  of  its  points ; 
and  an  angle  when  the  two  sides,  or  the  vertex  and  another  point  in 
each  of  the  sides.  The  tracing,  then,  of  a  system  of  lines,  circles, 
augles,  and  points,  aud,  consequently,  the  solution  of  a  problem  of 
Geometry,  when  this  problem  is  resolvable  by  the  aid  of  the  rule  and 
compass,  can  be  reduced  to  the  determination  of  a  certain  number 
of  unknown  points. 

W  c  may  call  thai  a  simple  problem  which  is  reduced  to  the  deter- 
mination of  a  single  unknown  point,  and  that  a  compound  problem 
which  requires  the  determination  of  several  points.  For  a  compound 
problem,  the  nature  of  the  solution  may  vary  not  only  with  the  num- 
ber and  nature  of  the  points  proposed  to  be  determined,  but  also 
with  the  order  of  their  determination  ;  and  it  is  easy  to  perceive  from 
hence  how  the  same  problem  of  Geometry  admits  of  different  solu- 
tions more  or  less  elegant.  But  as  the  different  unknown  points 
must  be  determined  one  after  another,  it  is  clear  that,  to  resolve  a 
compound  problem,  it  is  only  necessary  to  resolve  successively  a 
■amber  of  simple  problems. 

It  remains  to  consider  how  a  simple  problem  is  to  be  solved. 

In  every  simple  problem  the  unknown  point  is  generally  deter- 
mined by  two  conditions.  By  virtue  of  one  of  these  conditions  alone 
the  unknown  point  is  not  completely  determined  ;  it  will  only  be  sub- 
jected to  the  necessity  of  coinciding  with  one  of  the  points  situated 
upon  a  certain  right  line  or  curve  corresponding  to  this  condition. 
But  if  we  have  regard  to  the  conditions  united,  the  unknown  point 
must  be  situated,  at  the  same  time,  upon  the  two  lines  corresponding 
to  the  two  conditions,  and  can,  therefore,  only  be  at  one  of  the  points 
common  to  these  two  lines.  Then,  if  the  two  lines  do  not  meet,  the 
proposed  geometrical  problem  is  impossible,  or  incapable  of  solution. 
It  will  admit  of  a  single  solution,  if  the  two  lines  meet  in  a  single 
point ;  it  will  admit  of  several  distinct  solutions,  if  the  two  lines  in- 
tersect in  several  points.  Thus  a  simple  determinate  problem  may 
be  considered  as  resulting  from  the  combination  of  two  other  simple 
problems,  but  indeterminate,  each  of  which  consists  in  finding  a  point 
which  fp'.llls  a  single  condition,  or,  rather,  the  geometric  locus  of  all 
(he  pi'its  (infinite  in  number)  which  fulfill  the  given  condition.  If 
this  Condition  is  reduced  to  that  of  the  unknown  point  being  fond 
upon  a  certain  line,  the  geometric  locus  sought  will  evidently  be  this 
line  itself.  It  may  be  added,  that  very  often  the  geometric  locus  cor- 
responding to  a  given  condition  will  comprehend  the  system  of  a 
number  of  right  lines  or  curves.     Thu.--f  for  instance,  if  the  unknown 


94  GEOMETRY. 

point  is  required  to  be  at  a  given  distance  from  a  given  right  line,  the 
geometric  locus  sought  will  be  the  system  of  two  parallel  lines  drawn 
at  the  given  distance  from  this  line. 

Let  it  be  observed,  moreover,  that  a  simple  problem,  determinate 
or  indeterminate,  will  be  resolvable  by  the  rule  and  dividers,  if  each 
of  the  geometric  loci  which  serve  to  resolve  it  is  reduced  to  a  system 
of  right  lines  or  circumferences. 

To  illustrate  the  above,  the  solutions  of  some  simple  and  indeterm- 
inate problems  will  now  be  indicated. 

1°  Prob.  To  find  a  point  which  shall  be  situated  upon  a  given  line. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
line  itself. 

2°  Prob.  To  find  a  point  which  shall  be  situate  upon  the  circumfe- 
rence of  a  given  circle. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
circumference  of  the  given  circle  itself. 

3°  Prob.  To  find  a  point  which  shall  be  at  a  given  distance  from  a 
given  pohit. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
circumference  of  a  circle  described  with  the  given  point  as  a  center, 
and  with  a  radius  equal  to  the  given  distance. 

4°  Prob.  To  find  a  point  which  shall  be  situated  at  a  given  distance 
from  a  given  line. 

Solution.  The  geometric  locus  wrhich  resolves  this  problem  is  the 
system  of  two  lines  drawn  parallel  to  the  given  line,  and  separated 
from  it  by  the  given  distance. 

5°  Prob.  To  find  a  point  which  shall  be  at  a  given  distance  from 
the  circumference  of  a  given  circle. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
system  of  two  circumferences  of  circles  which  are  concentric  with 
the  given  circle,  and  have  radii  equal  to  its  radius,  increased  or  dimin- 
ished by  the  given  distance. 

6°  Prob.  To  find  a  point  which  shall  be  situated  at  equal  distances 
from  two  given  points. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
perpendicular  erected  at  the  middle  of  the  line  which  joins  the  two 
given  points. 

7°  Prob.  To  find  a  point  which  shall  be  situated  at  equal  distances 
from  two  given  parallel  lines. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a 
third  line  parallel  to  the  two  others,  and  which  divides  their  mutual 
distance  into  two  equal  parts. 

8°  Prob.  To  find  a  point  which  shall  be  at  equal  distances  from  two 
fines  which  intersect. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
system  of  two  new  lines  which  bisect  the  angles  comprehended  be- 
tween the  given  lines. 

9°  Prob.  To  find  a  point  situated  at  equal  distances  from  the  cir- 
cumferences of  two  given  concentric  circles. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a 
third  circumference  concentric  to  the  other  two,  and  which  divides 
their  mutual  distance  into  equal  parts. 

10°  Prob.  To  find  a  point  from  which  lines  drawn  to  the  extrem- 


PROBLEMS.  95 

!  ;i  line  given  in  length  and  position,  form,  with  each  other,  a 
right  angle. 

Solution.  The  geometric  locus  which  resolves  this  problem  ■  tho 
circumference  of  a  circle  which  has  the  given  line  for  a  diameter. 

11°  Prob.  To  find  a  }xrint  from  which  lines  drawn  to  the  extremi- 
ties of  a  given  line  form,  with  each  other,  an  obtuse  or  acute  angle. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
system  of  two  segments  of  a  circle  described  on  the  given  line  as  a 
coord,  and  capable  of  containing  the  given  angle. 

12°  Prob.  To  find  a  point  the  distances  of  which,  from  two  given 
points,  shall  have  a  given  ratio. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
circumference  of  a  circle,  one  diameter  of  which  has,  lor  extremities, 
the  two  points  which  fulfill  the  prescribed  condition  upon  the  line 
drawn  through  the  two  given  points. 

13°  Prob.  To  find  a  point  the  distances  of  which,  from  two  given 
lines,  6hall  be  in  a  given  ratio. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
■g -stem  of  two  new  fines  which  divide  the  angles  comprehended  be- 
tween the  given  lines  into  parts,  the  trigonometric  sines  of  which 
have  the  given  ratio.* 

14°  Prob.  To  find  a  point  the  distances  of  which,  from  two  given 
points,  are  the  sides  of  squares,  the  difference  of  which  is  equal  to  a 
given  square. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
perpendicular  erected  upon  the  line  which  joins  the  two  given  points 
at  the  point  of  this  line  which  fulfills  the  given  condition. 

15°  Prob.  To  find  a  point  the  distances  of  which,  from  two  given 
points,  are  sides  of  squares,  the  sum  of  which  is  equal  to  a  given 
square. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
circumference  of  a  circle,  one  diameter  of  which  has,  lor  extremities, 
the  two  points  which  fulfill  the  prescribed  condition,  upon  the  line 
joining  the  two  given  points. 

16°  Prob.  To  hud  a  point  such  that  the  oblique  line  drawn  from 
this  point  to  a  given  line,  under  a  given  angle,  shall  have  a  given 
length. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a  sys- 
tem of  two  lines  drawn  parallel  to  the  given  line  through  the  extrem- 
ities of  a  secant  line,  which,  having  its  middle  point  upon  the  given 
line,  cuts  it  at  the  given  angle,  aud  has  a  length  double  the  given 
length. 

17°  Prob.  To  find  a  point  such  that  the  secant,  drawn  from  this 
point  to  the  circumference  of  a  given  circle  and  parallel  to  a  given 
liii'".  shall  be  of  given  length. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
system  of  two  new  circumferences,  the  radii  of  which  are  equal  to 
that  of  the  given  circumference,  and  the  centers  of  which  are  the  ex- 
tremities of  a  line  which,  having  its  middle  point  at  the  center  of  the 
given  circle,  is  parallel  to  the  given  fine,  and  of  a  length  equal  to  doub- 
le the  given  length. 

*  This  solution,  of  course,  requires  a  knowledge  of  the  first  princi- 
ples of  Trigonometry. 


96  GEOMETRY. 

18°  Prob.  A  point  and  a  line  being  given,  to  find  a  second  point 
which  shall  be  the  middle  of  a  secant  drawn  from  the  given  point  to 
the  given  line. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a 
new  line  drawn  parallel  to  the  given  line,  and  which  divides  into 
ecpial  parts  the  distance  of  the  given  point  from  this  line. 

19°  Prob.  A  point  and  the  circumference  of  a  circle  being  given,  to 
find  a  second  point  which  shall  be  the  middle  of  a  secant  drawn  from 
this  point  to  the  circumference. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a 
new  circumference  of  a  circle  which  has  for  its  radius  the  half  of  ths 
radius  of  the  given  circumference,  and  for  its  center  the  middle  of 
the  distance  of  the  given  point,  from  the  center  of  the  given  circle. 

20°  Prob.  To  find  a  point  the  distance  of  which,  from  a  given 
point,  has  its  middle  upon  a  given  line. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a  new 
line  drawn  parallel  to  the  given  line,  at  a  distance  equal  to  that  which 
separates  this  line  from  the  given  point. 

21°  Prob.  To  find  a  point  the  distance  of  which,  from  a  given 
point,  has  its  middle  upon  the  circumference  of  a  given  circle. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a  new 
circumference  which  has  for  its  radius  the  double  of  the  radius  of  the 
given  circumference,  and  for  its  center  the  extremity  of  a  line,  the 
half  of  which  is  the  distance  of  the  given  point  from  the  center  of  the 
given  circle. 

22°  Prob.  Two  points  being  given,  symmetrically  placed  on  oppo- 
site sides  of  a  given  axis,  to  find  a  third  point  such  that  the  line  drawn 
from  this  third  point  to  the  first  shall  meet  the  given  axis  at  equal  dis- 
tances from  the  second  and  third  points. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a  line 
drawn  parallel  to  the  given  axis,  at  a  distance  equal  to  that  which  sep- 
arates it  from  the  given  point. 

23°  Prob.  A  circle  being  given  and  a  chord,  to  find  a  point  such 
that  the  line  drawn  from  this  point  to  one  of  the  extremities  of  a  chord 
shall  meet  the  circumference  of  the  circle  at  equal  distances  from  this 
point  and  from  the  other  extremity. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
system  of  two  new  circumferences  of  circles  which  have  for  a  common 
chord  the  given  chord,  and  for  centers  the  extremities  of  the  diame- 
ter perpendicular  to  this  chord  in  the  given  circle. 

24°  Prob.  Two  lines  perpendicular  to  each  other  being  given,  to 
find  a  point  which  shall  be  at  the  middle  of  a  secant  of  given  length 
comprehended  between  these  two  lines. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a  cir- 
cumference of  a  circle  which  has  for  its  center  the  point  common  to 
the  two  lines,  and  for  its  radius  half  the  given  length. 

25°  Prob.  To  find  in  a  given  circle  a  point  which  shall  be  the  mid- 
dle of  a  chord  of  given  length. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a  cir- 
cumference wrhich  has  for  its  center  the  center  of  the  given  circle, 
and  for  its  radius  the  distance  from  this  center  to  any  one  of  the 
chords  drawn  so  as  to  be  of  the  given  length. 

26°  Prob.  To  find  out  of  a  given  circle  a  point  which  must  be  the 
extremity  of  a  tangent  of  given  length. 


PROBLEMS.  97 

Solution.  The  geometric  locus  which  resolves  this  problem  is  a  cir- 
cumference  of  a  circle  which  has  lor  its  center  the  center  of  the  gives 
circle,  and  for  its  radius  the  distance  from  this  center  to  the  extremi- 
ty of  any  one  whatever  of  the  tangents,  drawn  in  such  a  manner  as  to 
be  of  the  given  length. 

27°  Prob.  To  find  out  of  a  given  circle  the  point  of  meeting  of 
two  tangents,  drawn  through  the  extremities  of  a  chord  which  con- 
tains a  given  point. 

Solution.  The  geometric  locus  which  resolves  this  problem  is  the 
polar  line  corresponding  to  the  given  point.     (See  Appendix  II.) 

The  solutions  above  given  are  easily  deduced  from  well-known  the- 
orems of  Geometry.  A  great  number  of  problems,  both  simple  and 
indeterminate,  could  be  pointed  out,  the  solutions  of  which  would  re- 
duce themselves,  m  a  similar  manner,  to  systems  of  right  lines  and, 
circumferences  of  circles.  Let  it  be  observed,  moreover,  that  from 
the  solutions  of  n  problem*  of  this  kind,  in  each  of  which  the  unknown 
point  is  subjected  to  a  single  condition,  we  can  deduce  immediately 

the  solutions  of  '        simple  and  determinate  problems,  in  each  of 

which  the  unknown  point  is  subjected  to  two  conditions.  For,  to  ob- 
tain a  simple  and  determinate  problem,  it  is  sufficient  to  combine  two 
conditions  corresponding  to  two  simple  but  indeterminate  problems, 
or  even  two  conditions  alike  and  corresponding  to  a  single  indeterm- 
inate problem.  But  the  number  of  combinations  of  n  quantities,  two 
and  two,  is  (see  Alg.,  art.  203), 

n(n— 1). 
2 
and.  adding  to  this  number  that  of  the  quantities  themselves,  the  re- 
sult is, 

n(n — 1)       _n(n-\-l) 
2  2 

This  result  increases  very  rapidly  with  n.     Thus,  if  n=27,       t7~   ■ 

=378;  that  is.  the  solution  of  the  27  indeterminate  problems  enun- 
ciated above  furnishes  already  the  means  of  resolving  378  simpler 
and  determinate  problems. 

In  order  that  the  principles  just  brought  to  view  maybe  the  better 
apprehended,  they  will  now  be  applied  to  the  solution  of  some  de- 
terminate problems. 

Suppose,  first,  that  it  is  required  to  draw  a  tangent  to  a  circle 
through  a  point  without.  The  question  may  be  reduced  to  seeking 
the  unknown  point  of  contact  Of  the  tangent  with  the  circle.  The 
two  conditions  which  this  point  must  satisfy  are,  1°.   That  it  shall  be 

upon  the  circumference  of  the  given  circle.    2~.  That  lines  drawn 

from  this  point  to  the  given  point  and  the  center  of  the  circle  should 
make  a  right  angle  with  each  other.  Then  the  question  to  be  re- 
solved will  be  a  determinate  problem,  resulting  from  the  combina- 
tions of  the  indeterminate  problems,  -2  and  10. 

The  combined  solutions  of  9  and  10  furnish,  in  fact,  the  solutions 
heretofore  given  (Prob.  20,  p.  83). 

Suppose,  secondly,  that  it  ■  required  to  circumscribe  a  circle  about 
The  question  can  be  reduced  to  seeking  for  the  cen- 

E 


'JO  GEOMETRY. 

ter  of  the  circle.  But  the  two  conditions  which  this  center  nr. 
isfy  will  be  those  <>f  being  not  only  at  equal  distances  from  the  first 
and  second  vertex  of  the  given  triangle,  but  also  at  equal  distances 
from  the  first  and  third.  Then  the  question  to  resolve  will  be  a  de- 
terminate problem  resulting  from  the  combination  of  two  indeterm- 
inate problems  identical  with  each  other  and  with  problem  G.  In 
fact,  the  solution  of  problem  6,  twice  repeated,  will  furnish  two  geo- 
metric loci,  reduced  to  two  right  lines,  which  cut  each  other  in  a  sin- 
gle point,  and  thus  will  be  obtained  the  known  solution  of  the  prob- 
lem proposed. 

Suppose,  next,  that  the  question  is  how  to  draw  a  circle  tangent  to 
the  three  sides  of  a  given  triangle. 

The  question  can  be  reduced  to  finding  the  center  of  the  given  cir- 
cle. But  the  two  conditions  which  this  center  must  satisfy  will  be 
not  only  to  be  at  equal  distances  from  the  first  and  second  side  of  the 
given  triangle,  but  also  at  equal  distances  from  the  first  and  third 
sides.  Then  the  question  to  be  resolved  will  be  a  determinate  prob- 
lem, resulting  from  the  combination  of  two  indeterminate  problems 
identical  with  one  another,  and  with  problem  8.  In  fact,  the  solution 
of  problem  8  twice  will  furnish  two  geometric  loci,  which,  reduced 
each  to  the  system  of  two  right  lines,  will  cut  each  other  in  four  points, 
and  thus  four  solutions  will  be  obtained  of  the  proposed  problem. 

Suppose,  finally,  that  the  question  is  to  inscribe  between  a  chord 
of  a  circle  and  its  circumference  a  line  equal  and  parallel  to  a  given 
line.  The  question  can  be  reduced  to  seeking  either  one  of  the  two 
points  which  will  form  the  extremities  of  this  line,  and,  consequently, 
to  a  determinate  problem  resulting  from  the  combination  of  two  hide- 
terminate  problems,  to  wit,  problems  1  and  17,  or  problems  2  and 
16.  In  fact,  by  the  aid  of  this  combination,  the  question  proposed  is 
resolved  without  difficulty.  And  one  of  the  extremities  of  the  line 
sought  will  be  found  determined  either  by  the  meeting  of  the  circum- 
ference of  the  given  circle  with  a  new  line,  or  by  the  meeting  of  the 
given  chord  with  a  new  circumference. 

It  is  seen  here  how  the  solution  obtained  may  be  modified  when 
the  order  comes  to  be  inverted  in  which  the  unknown  points  are  de- 
termined. 

The  construction  of  the  geometric  locus  which  corresponds  to  a 
simple  and  indeterminate  problem  may  itself  require  the  resolution 
of  one  or  more  determinate  problems.  It  should  be  observed,  upon 
this  subject,  that  in  the  case  where  the  problem  is  resolvable  by  the 
rule  and  compass,  the  geometric  locus  should  reduce  to  a  system  of 
right  lines  and  circles.  Then,  since  each  line  or  each  circumference 
finds  itself  completely  determined  when  there  are  known  two  or  three 
points  of  it,  the  construction  of  the  geometric  locus,  corresponding  to 
a  simple  and  indeterminate  problem,  can  always  be  deduced  from  the 
construction  of  a  certain  number  of  points  suitable  to  verify  the  con- 
dition which  ought  to  be  fulfilled,  in  virtue  of  the  enunciation  of  the 
problem,  by  the  unknown  point. 

Thus,  for  example,  to  resolve  problem  6  ;  that  is  to  say,  to  find  a 
point  which  shall  be  situated  at  equal  distances  from  two  given  points, 
and,  consequently,  to  construct  the  geometric  locus  which  shall  con- 
tain every  point  suitable  to  fulfill  this  condition.  We  begin  by  seek- 
ing such  a  point,  for  example,  one  the  distance  of  which  from  the 


PROBLEMS.  99 

given  points  is  sufficiently  great.  But  the  solution  of  this  last  prob- 
lem deduces  itself  immediately  from  problem  'h  which,  twice  repeat- 
ed, will  furnish  ;it  once  two  points  that  fulfill  the  proposed  condition; 
consequently,  two  points  which  suffice  to  determine  the  geometric 
locus  required. 

Tims,  again,  to  resolve  problem  15;  that  is  to  say,  to  find  a  point 
the  distances  of  which,  from  two  points  given,  shall  furnish  squares 
the  sum  of  which  shall  be  equal  to  a  given  square,  and,  consequently, 
to  construct  the  geometric  locus  of  every  point  suitable  to  fulfill  this 
condition.  We  can  commence  by  seeking  such  a  point;  for  exam- 
ple, that  which  shall  be  situated  at  equal  distances  from  the  two  given 
points,  and,  consequently,  separated  from  each  of  them  by  a  distance 
ecjual  to  half  the  diagonal  of  the  given  square.  But  the  solution  of 
this  last  problem  deduces  itself  immediately  from  the  solution  of  prob- 
lem 3,  and,  twice  repeated,  will  furnish  at  once,  also,  two  points  which 
will  fulfil]  the  proposed  condition.  Moreover,  these  two  points  are 
precisely  the  extremities  of  a  diameter  of  the  circle,  the  circumference 
of  which  represents  the  geometric  locus  required. 

The  above  note  is  from  a  recent  article  by  the  celebrated  Cauchy. 
Although  designed  by  him  as  an  introduction  to  a  new  method  of  re- 
solving determinate  geometric  problems  by  means  of  the  Indetermin- 
ate Analysis  (for  an  exhibition  of  which,  see  Comptes  Rendus  de 
L'Acadamie  des  Sciences,  No.  17,  21  Avril,  1843,  p.  8G7,  and  No.  19, 
IS  Mai,  1843,  p.  1039),  yet  it  is  calculated  to  afford  important  aid  to 
the  solution  of  problems  by  the  processes  of  ordinary  geometry. 

M.  Cauchy  acknowledges  that  it  is  but  the  development  Of  some 
principles,  the  memory  of  which  he  has  preserved,  which  were  con- 
tained in  the  course  of  lectures  given  by  Dinet,  at  the  Lycee  Napoleon, 
some  forty  years  ago. 


100  GEOMETRY. 


MISCELLANEOUS  EXERCISES  IN  PLANE  GEOMETRY. 

1.  Prove  that  all  regular  polygons  of  the  same  number  of  sides  are 
similar  figures. 

2.  That  if  a  line  join  the  middle  points  of  two  sides  of  a  triangle, 
it  will  be  parallel  to  the  third  side  and  equal  to  its  half. 

3.  To  describe  a  circle  about  a  given  square. 

4.  To  divide  a  right  angle  into  three  equal  parts. 

5.  To  circumscribe  about  a  given  circle  a  triangle  one  side  of 
which  is  given. 

6.  Find  the  length  of  the  circumference  of  a  circle  in  seconds  of  a 
degree. 

7.  Find  the  length  of  the  radius  in  seconds  of  a  degree. 

8.  Find  the  length  of  1"  to  radius  1. 

9.  Prove  that  a  straight  line  can  meet  a  circumference  in  but  two 
points. 

10.  From  a  given  point  without  a  circle  to  draw  a  secant  such 
that  the  part  within  the  circle  shall  be  equal  to  a  given  line. 

11.  To  draw  to  a  circle  a  tangent  of  given  length,  and  termina- 
ting at  a  given  line,  which  cuts  the  circumference. 

12.  An  inscribed  polygon  being  given,  to  circumscribe  another 
similar. 

13.  Prove  that  of  two  convex  lines,  broken  or  curved,  termina- 
ting at  the  same  points,  the  enveloped  is  less  than  the  enveloping  line. 

14.  Prove  that  the  difference  between  the  sum  of  the  two  per- 
pendicular sides  of  a  right-angled  triangle  and  the  hypothenuse  is 
equal  to  the  diameter  of  the  inscribed  circle.  , 

15.  To  trisect  a  given  finite  straight  line. 

16.  Prove  that  if,  from  the  extremities  of  the  diameter  of  a  sem- 
icircle, perpendiculars  be  let  fall  on  any  line  cutting  the  semicircle, 
the  parts  intercepted  between  those  perpendiculars  and  the  circum- 
ference are  equal. 

17.  If,  on  each  side  of  any  point  in  a  circle,  any  number  of  equal 
arcs  be  taken,  and  the  extremities  of  each  pair  joined,  the  sum  of  the 
chords  so  drawn  will  be  equal  to  the  last  chord  produced  to  meet  a 
line  drawn  from  the  given  point  through  the  extremity  of  the  first  arc. 

18.  That  if  two  circles  touch  each  other,  and  also  touch  a  straight 
line,  the  part  of  the  line  between  the  points  of  contact  is  a  mean  pro- 
portional between  the  diameters  of  the  circles, 

19.  From  two  given  points  in  the  circumference  of  a  given  circle 
to  draw  two  lines  to  a  point  in  the  same  circumference,  which  6hall 


MISCELLANEOUS    EXERCI.i;:  ■=.  lOi 

cut  a  line  giveu  in  position,  so  that  the  part  of  it  intercepted  hy  them 
may  be  equal  to  a  given  line. 

20.  Prove  that  if,  from  any  point  within  an  equilateral  triangle, 
perpendiculars  be  drawn  to  the  sides,  they  are  together  equal  to  a 
perpendicular  drawn  from  any  of  the  angles  to  the  opposite  side. 

21.  That  if  the  three  sides  of  a  triangle  be  bisected,  the  perpen- 
diculars drawn  to  the  sides,  at  the  three  points  of  bisection,  will  meet 
in  the  same  point. 

22.  If  from  the  three  vertices  of  a  triangle  lines  be  drawn  to  the 
points  of  bisection  of  the  opposite  sides,  these  Hues  intersect  each 
other  in  the  same  point. 

23.  The  three  straight  lines  which  bisect  the  three  angles  of  a 
triangle  meet  in  the  same  point. 

24.  If  from  the  angles  of  a  triangle  perpendiculars  be  drawn  to 
the  opposite  sides,  they  will  intersect  in  the  same  point. 

•J  -'».  If  any  two  chords  be  drawn  in  a  circle,  to  intersect  at  right 
angles,  the  sum  of  the  squares  of  the  four  segments  is  equal  to  the 
square  of  the  diameter  of  the  circle. 

26.  In  a  given  triangle  to  inscribe  a  rectangle  whose  sides  shall 
have  a  given  ratio. 

27.  Prove  that  the  two  sides  of  a  triangle  are  together  greater 
than  the  double  of  the  straight  line  which  joins  the  vertex  and  the 
bisection  of  the  base. 

28.  That  if,  in  the  sides  of  a  square,  at  equal  distances  from  the 
four  angles,  four  other  points  be  taken,  one  in  each  side,  the  figure 
contained  by  the  straight  lines  which  join  them  shall  also  be  a  square. 

29.  That  if  the  sides  of  an  equilateral  and  equiangular  pentagon 
be  produced  to  meet,  the  angles  formed  by  these  lines  are  together 
equal  to  two  right  angles. 

30.  That  if  the  sides  of  an  equilateral  and  equiangular  hexagon 
be  produced  to  meet,  the  angles  formed  by  these  lines  are  together 
equal  to  four  right  angles. 

31.  If  squares  be  described  on  the  three  sides  of  a  right-angled 
triangle,  and  the  extremities  of  the  adjacent  sides  of  any  two  squares 
be  joined,  the  triangles  so  formed  are  equal,  though  not  identical, 
to  the  giveu  triangle,  and  to  one  another. 

32.  If  the  squares  be  described  on  the  hypothenuse  and  sides  of 
a  right-angled  triangle,  and  the  extremities  of  the  sides  of  the  former 
square,  and  thoso  of  the  adjacent  sides  of  the  others,  be  joined,  the 
sum  of  the  squares  of  the  lines  joining  them  will  be  equal  to  five 
times  the  square  of  the  hypothenuse. 

33.  To  bisect  a  triangle  by  a  line  drawn  parallel  to  one  of  its  sides. 


JQ2  GEOMETRY'. 


34.  To  divide  a  circle  into  any  number  of  concentric  equal  aunuli. 

35.  To  inscribe  a  square  in  a  given  semicircle. 

36.  Prove  that  if,  on  one  side  of  an  equilateral  triangle,  as  a  diam- 
eter, a  semicircle  be  described,  and  from  the  opposite  angle  two 
straight  lines  be  drawn  to  trisect  that  side,  these  lines  produced  will 
trisect  the  semi-circumference. 

37.  Draw  straight  lines  across  the  angles  of  a  given  square,  so  as  to 
form  an  equilateral  and  equiangular  octagon. 

38.  Prove  that  the  square  of  the  side  of  an  equilateral  triangle,  in- 
scribed in  a  circle,  is  equal  to  three  times  the  square  of  the  radius. 

39.  To  draw  straight  lines  from  the  extremities  of  a  chord  to  a  point 
in  the  circumference  of  the  circle,  so  that  their  sum  shall  be  equal  to 
a  given  line.     N.B.  The  given  line  must  evidently  be  limited. 

40.  In  a  given  triangle  to  inscribe  a  rectangle  of  a  given  area. 

41.  Given  the  perimeter  of  a  right-angled  triangle,  and  the  perpen- 
dicular from  the  right  angle  upon  the  hypothenuse,  to  construct  the 
triangle. 

42.  In  an  isosceles  triangle  to  inscribe  three  circles  touching  each 
other,  and  each  touching  two  of  the  three  sides  of  the  triangle. 

43.  To  construct  a  trapezoid  when  four  sides  are  given. 

44.  The  same  when  three  sides  and  the  sum  of  the  angles  at  the 
base  are  given. 

45.  When  the  two  sides  not  paralel,  the  altitude  and  an  angle 
are  given. 

46.  When  the  difference  of  the  parallel  sides,  the  diagonal,  a  third 
side,  and  an  augle. 

47.  Prove  that  the  line  drawn  to  the  middle  of  the  hypothenuse 
from  the  vertex  of  the  right  angle  in  a  right-angled  triangle  is  equal 
to  half  the  hypothenuse. 

48.  Prove  that  if  the  four  angles  of  a  parallelogram  be  bisected, 
and  the  points  in  which  two  of  the  bisecting  lines  adjacent  one  side 
meet  be  joined  with  that  in  which  the  two  bisecting  lines  adjacent 
the  opposite  side  meet;  1°,  that  the  joining  line  will  be  parallel  to  the 
other  two  sides;  2°,  that  it  will  be  equal  to  the  difference  of  two 
adjacent  sides. 

49.  That  in  any  quadrilateral  the  lines  joining  the  middle  points 
of  the  opposite  sides,  and  the  line  joining  the  middle  points  of  the 
diagonals,  meet  in  the  same  point,  and  all  three  bisect  one  another. 

50.  That  the  rectangle  of  the  two  sides  of  any  triangle  is  equal  to 
the  rectangle  of  the  perpendicular  upon  the  third  side  from  the  ver- 
tex opposite, 
angle. 


MISCELLANEOUS    EXERCISES.  103 

51.  Prove  that  the  rectangle  ql  the  diagonals  of  an  inscribed  quad- 
rilateral is  equal  to  the  sum  of  the  rectangles  of  the  opposite  sides. 

52.  Prove  that  the  square  of  the  line  bisecting  the  vertical  angle  of 
a  triangle,  together  with  the  rectangle  of  the  two  segments  of  the 
baae,  is  equal  to  the  rectangle  of  the  other  two  sides  of  the  triangle. 

53.  To  find  two  lines  that  shall  have  the  same  ratio  as  two  given 
rectangles. 

54.  Draw  a  transverse  line  to  two  circles  such  that  the  parts  com- 
prehended  within  the  circumferences  shall  be  equal  to  a  given  line. 

55.  To  inscribe  in  a  circle  (radius  not  given)  a  triangle  of  given 
base,  vertical  angle,  and  altitude. 

56.  In  a  given  circle  to  place  six  others,  so  that  each  shall  touch 
two  others,  and  the  given. 

57.  To  construct  a  figure  similar  to  two  given  similar  figures,  an  J 
equal  to  their  sum  or  difference. 


APPENDIX  I. 


ISOPERIMETRY. 

Def  1.  A  maximum  is  the  greatest  quantity  among  those  of  the 
same  kind  ;  a  minimum  the  least.* 

Def.  2.  Isoperimetrical  figures  are  those  which  have  equal  periui- 
MttB. 

THEOREM    I. 

Among  all  triangles  of  the  same  perimeter,  that  is,  a  maximum  in 
which  the  undetermined  sides  are  equal. 

Let  ABC,  ABD  be  the  two  triangles. 
With  C  as  center,  and  radius  CA,  de- 
scribe a  circle  cutting  AC  produced  in 
F ;  ABF,  inscribed  in  a  semicircle,  will 
be  a  right  angle ;  produce  FB,  making 
DE  =  DB,  and  draw  the  perpendiculars 
DII,  CG.  It  will  be  easy  to  show  of 
the  oblique  lines  that  AE  <  AF  .-.  BE 
<  BF  .\  BH  <  BG;  these  last  two  be- 
ing the  altitudes  of  the  given  triangles 
which  have  a  common  base. 


THEOREM    II. 

Of  all  isoperimetric  polygons  the  maximum  has  its  sides  equal. 
By  drawing  a  diagonal  in  the  polygon  so  as  to  cut  off  two  sides 
forming  a  triangle  ;  if  these  two  sides  be  not  equal,  they  may  be  re- 
placed fcy  two  others  which  are  equal,  and  which,  by  the  last  theorem, 
will  inclose  a  greater  triangle.  This  process  may  be  repeated  with 
all  the  sides  of  the  polygon. 

THEOREM    III. 

Of  all  triangles  formed  with  two  given  sides,  that  is,  a  maximum  in 
which  the  two  given  sides  make  a  right  angle. 

For  with  the  same  base  it  will  have  the  greater  altitude. 


*  These  definitions  are  suited  to  our  present  purpose. 


GEOMETRY. 


THEOREM    IV. 

Of  all  polygons  formed  with  sides  all  given  except  one  side,  that  is, 
a  maximum,  of  which  the  given  sides  are  inscribed  in  a  semicircle,  of 
which  the  side  not  given  is  the  diameter. 

Let  ABCDEF  be  the  maxim- 
um polygon  formed  with  sides 
all  given  except  AB.  Join  AD, 
BD  ;  then  ADB  must  be  a  right 
angle :  otherwise,  preserving  the 
parts  BCD  and  ADEF  the  same,  Ai 
the  triangle  ADB  might  be  increased  (th.  3)  ;  the  point  D  must,  there- 
fore, be  in  the  semicircumference  described  on  AB  as  a  diameter. 
In  the  same  manner  it  may  be  proved  that  the  points  E,  F,  C,  &c., 
must  be  in  the  semicircumference.     Q.  E.  D. 

Schol.  There  is  but  one  way  of  forming  the  polygon ;  for  if,  after 
having  found  a  circle  which  satisfies  the  requisition,  a  larger  circle  be 
supposed,  the  chords  which  are  the  sides  of  the  polygon  correspond 
to  smaller  angles  at  the  center,  and  the  sum  of  these  will  be  less  than 
two  right  angles. 


THEOREM    V. 


Of  all  polygons  formed  with  given  sides,  thai  is,  a  maximum  which 
can  be  inscribed  in  a  circle. 


Let  ABCEFG  be  an  inscribed  polygon,  abcefg  one  which  is  not 
capable  of  being  inscribed,  and  of  equal  sides  with  the  former ;  draw 
the  diameter  EM  ;  join  AM,  MB;  upon  ab  make  the  triangle  abm  = 
ABM,  and  join  em.  By  th.  4,  EFGAM  >  efgam  and  ECBM>  ecbm 
.-.  by  addition,  EFGAMBC  minus  AMB  >  efgambc  minus  amb.  Q. 
E.D. 

Corollary  from  the  two  last  propositions.  The  regular  polygon  is 
the  maximum  among  all  isoperimetric  polygons  of  the  same  number  of 
sides. 


APPENDIX  I. 


TMEORKM    VI. 


Of  all  regular  isoperimelric  polygons,  that  is  the  greatest  which  has 
the  greatest  number  of  sides. 


Let  DE  be  the  half  side  of  one  of  the  polygons,  O  its  center,  OE  its 
apothegm.  AB,  C,  CB  the  same  for  the  other;  DOE,  ACB  will  be 
the  half  angles  at  the  centers  of  the  polygons ;  and  as  these  angles  are 
not  equal,  the  lines  C  A,  OD,  prolonged,  will  meet  at  the  point  F ;  from 
this  point  draw  FG  perpendicular  to  OC  ;  with  O  and  C  as  centers, 
describe  arcs  GI,  GH. 

Now,  DE  is  to  the  perimeter  of  the  first  polygon  as  O  is  to  four  right 
angles,  and  AB  :  perim.  2d  polyg. : :  C  :  4  r.  angs. .-.  DE  :  AB  : :  O  :  C, 

and  .-.  (th.  71,  corol.  3),  DE  :  AB  :  :^L:  Cl!l.     Multiplying  the  ante 

OG    CG 

cedents  by  OG,  and  the  consequents  by  CG, 

DE  X  OG:AB  XCG::GI:GH. 

But  the  similar  triangles  ODE,  OFG  give 

OE  :  OG  : :  DE  :  FG  .-.  OE  X  FG  =  DE  X  OG  (th.  54). 

In  the  same  manner  it  may  be  shown  that 

AB  X  CG  =  CB  X  FG 

Therefore,  by  substitution, 

OE  X  FG :  CB  X  FG  :  GI :  GH. 

If,  then,  it  can  be  shown  that  the  arc  GI  >  arc  GH,  it  will  follow  that 

the  apothegm  OE  is  greater  than  the  apothegm  CB. 

Make  the  figures    CKx  =  CGx,  CKH  =  CGH. 

Then  KcG  >  KHG  (see  exercise  13  of  the  miscellaneous  exercises). 


4  GEOMETRY. 

.-.  Gz=  I  K*G  >  GH  =  iKHG. 
Much  more  GI  >  GH.  Q.  E.  D.* 

Corollary  from  the  preceding  Propositions. — The  circle  is  the  great- 
est of  all  figures  of  the  same  perimeter;  for  it  may  be  regarded  as  a 
regular  polygon  of  an  infinite  number  of  sides. 

*  It  has  been  seen  in  the  note  to  corol.  4,  th.  1G,  that  equilateral 
triangles,  squares,  and  regular  hexagons  are  the  only  figures  which 
will,  in  juxtaposition,  leave  no  intervening  space.  It  appears,  also, 
from  the  present  proposition,  that  the  6pace  inclosed  in  the  regular 
hexagon  is  greater  than  that  inclosed  in  the  square  or  triangle  of  the 
same  perimeter.  Some  writers  on  natural  theology  call  attention  to 
the  fact  that  the  cells  of  the  bee-hive  being  made  in  the  form  of  reg- 
ular hexagons,  thus  affording  the  greatest  space  with  a  given  amount 
of  the  material  employed  in  their  construction,  indicate  an  instinct 
working  in  accordance  with  the  most  recondite  principles  of  geometry. 


APPENDIX  II. 


CENTERS  OF  SYMMETRY. 

Def.  1.  When  the  vertice8  of  two  polygons,  or  of  the  same  polygon, 
are  two  and  two  upon  lines  meeting  in  a  point  interior,  and  at  equal 
distances  from  this  point,  the  point  is  called  the  center  of  symmetry. 

Theorem  1.  Prove  that  all  lines  drawn  to  opposite  parts  of  the 
figure  through  the  center  of  symmetry  are  equally  divided  at  this 
point.  2.  That  the  opposite  sides  of  the  figure  or  figures  are  equal, 
parallel,  and  arranged  In  a  reverse  order.     3.  The  converse. 

Def.  2.  Two  points  are  said  he  situated  symmetrically  with  respect 
to  a  line  when  this  line  is  perpendicular  to  that  which  joins  the  two 
points,  and  divides  it  into  two  equal  parts. 

OF  AXES  OF  SYMMETRY. 

Two  polygons,  or  portions  of  one  polygon,  are  said  to  be  symmet- 
rical with  respect  to  a  line  when  their  corresponding  vertices  are 
symmetrical.     The  line  in  such  a  case  is  called  an  axis  of  symmetry. 

Def.  3.  An  isosceles  trapezoid  is  one  whose  inclined  sides  are 
equal. 

Theorem.  Prove  that  the  line  joining  the  middle  points  of  the  par- 
allel bases  of  such  a  figure  is  an  axis  of  symmetry. 

Theorem  2.  Prove  that  the  isosceles  trapezoid  may  be  inscribed  in 
a  circle. 

General  Theorem.  Prove  that  every  figure  which  has  two  axes  of 
symmetry  perpendicular  to  each  other  has  a  center  of  symmetry  at 
their  intersection. 

Schol.  Show  that  these  axes  divide  the  figure  into  four  equal  parts. 

OF  DIAMETERS. 
Theorem  3.  When  the  vertices  of  two  polygons  or  of  a  same  polygon 
are  two  and  two   upon  lines   parallel,  and  equally  divided  by  a 
mkdian  line,  this  median  line  bisects,  also,  every  other  line  parallel  to 
the  former,  and  terminating  at  the  sides  of  the  figure  or  figures. 

CENTER  OF  MEAN  DISTANCES. 
Def.  4.  If  the  middle  points  of  the  consecutive  sides  of  a  polygon 
be  joined,  a  new  polygon  will  be  formed  of  less  perimeter  an<i 


2  GEOMETRY. 

evidently,  than  the  perimeter  and  area  of  the  first.  Proceeding  in 
the  same  manner  with  the  second,  a  third  is  obtained,  still  smaller, 
and  so  on.  These  operations  being  continued  indefinitely,  the  result 
will  be  at  length  a  polygon,  infinitely  small,  which  may  be  regarded 
as  a  point.  This  point  is  called  the  center  of  mean  distances.  It  has 
a  remarkable  property.  The  distance  of  this  point  from  any  given- 
line  is  equal  to  the  quotient  of  the  sum  of  the  distances  of  all  the  vertices 
of  the  polygon  from  this  given  line,  divided  by  the  number  of  vertices. 
The  student  may  prove  this  by  proving  the  sum  of  the  distances  of 
the  vertices  of  the  second  polygon  equal  to  that  of  the  first,  and  so  on, 
till  the  polygons  are  reduced  to  a  point,  the  center  of  mean  distances. 

Cor.  From  this  will  follow  a  construction  for  determining  the  cen- 
ter of  mean  distances,  viz.,  determine  its  distance  from  two  given 
lines  by  dividing  the  sum  of  the  distances  by  the  number  of  vertices 
of  the  polygon,  and,  drawing  parallels  to  these  two  lines  at  the  dis- 
tances thus  determined,  these  parallels  will,  by  their  intersection,  de- 
termine the  point  required. 

Def.  5.  Polygons  are  said  to  be  inversely  similar  when  one  is  similar 
to  a  polygon  symmetric  with  the  other. 

CENTERS  OF  SIMILITUDE. 

Def.  6.  The  center  of  similitude  is  a  point  placed  in  such  a  manner 
with  reference  to  two  polygons,  directly  similar,  as  that,  if  a  line  be 
drawn  through  it  to  two  homologous  vertices  of  the  polygons,  the 
direction  of  these  vertices  from  the  point  shall  be  the  same,  and  the 
lines  proportional  to  the  homologous  sides. 

The  distances  from  this  point  to  the  homologous  vertices  are  called 
radii  of  similitude. 

If,  from  a  point  taken  at  pleasure,  lines  be  drawn  to  the  vertices  of 
a  polygon,  and  upon  these  lines  or  their  prolongations  parts  be  taken 
proportional  to  them,  the  points  thus  obtained  will  determine  a  new 
polygon  similar  to  the  given  polygon,  and  the  arbitrary  point  will  be 
the  center  of  similitude  of  the  two  polygons. 

The  center  of  similitude  may  be  either  external  or  internal  to  the 
two  polygons.     (See  diagram  of  note  to  th.  69  for  an  external  center.) 

Two  similar  polygons  which  have  their  sides  respectively  parallel, 
and  directed  the  same  or  contrary  ways,  have  in  the  first  case  an  ex- 
ternal, and  in  the  second  case  an  internal  center  of  similitude.  In 
the  latter  case  it  is  between  homologous  vertices. 

Theorem.  Prove  that  when  three  similar  polygons  have  their  sides 
respectively  parallel,  their  three  centers  of  similitudes  are  upon  the 
line. 


APPENDIX  II.  d 

THEOREM. 

Two  polygons  (directly^  similar,  situated  in  any  manner  upon  a 
plane,  have  always  a  common  homologous  point* 

By  this  is  to  be  understood  that  there  exists  in  the  plane  of  the  two 
polygons  a  point  such  that,  if  it  be  joined  with  the  vertices  of  the 
two  polygons,  the  homologous  lines  of  junction  will  have  the  ratio  of 
similitude  of  the  two  polygons^  and  that  the  angles  formed  by  these 
lines  are  equal  each  to  each. 
E  D 


— ->* 


Let  ABODE,  abcde  be  the  two  polygons,  and  N  the  point  in  which 
the.  two  sides  AB,  ab  meet.  Find  P,  the  center  of  a  circle  passing 
through  A,  a  and  N,  or  equally  distant  from  these  three  points,  and 
Q  ■  point  equally  distant  from  B,  b  and  N.  Join  PQ.  Then  find  the 
point  0  symmetric  to  N,  with  reference  to  the  line  PA.  O  will  be 
the  point  required. 

*  That  is,  a  point  from  which  homologous  lines,  drawn  to  the  vert- 
ices of  the  two  polygons,  will  have  the  ratio  of  similitude  of  the  pol- 
ygons, nii*l  form  with  each  oilier  equal  angles. 

This  theorem  is  due  to  M.  Chasles,  and  is  demonstrated  in  the  Bul- 
letin des  Sciences  Mathematique  of  Ferussac  for  1830. 

t  See  note  to  def.  67. 


4  GEOMETRY. 

For  APN,  APO  are  isosceles  triangles,  and  give  the  angle  PAN  = 
PNA,  and  angle  PAO  =  POA  .-.  NPA'  =  2PAN,  OP  A'  =  2PAO ; 
or,  by  addition,  NPO  =  2NAO,  i.  e.,  NPO  =  2BAO. 

Similarly,  the  two  triangles  aPN,  aPO  give 

NPO  bs  2NaO,  or  NPO  =  2baO  .-.  BAO  =  baO. 

Reasoning  upon  the  four  points  Q,  B,  b,  N  in  the  same  manner  as 
upon  the  four  points  P,  A,  a,  N,  it  may  be  proved  that  the  angles  ABO 
and  abO  are  equal.  Thus  the  triangles  OAB,  Oab  are  similar,  and 
give 

OA  :  Oa  :  :  OB  :  Ob  :  :  AB  :  ab. 

The  same  would  result  from  any  number  of  triangles. 

Scholium.  The  point  O  is  the  only  common  homologous  point  of  the 
two  polygons.  Every  line  passing  through  it  is  called  a  common  ho- 
mologous line,  and  conversely. 

The  center  of  similitude  of  two  polygons  is  their  common  homolo- 
gous point. 

There  is  another  mode  of  construction  which  seems  more  natural 
than  that  just  given. 

Determine  the  circumference  of  a  circle,  every  point  of  which  shall 
be  at  distances  from  two  homologous  vertices  A,  a  in  the  ratio  of  simil- 
itude of  the  two  polygons  (see  Prob.  12  of  General  Note,  p.  95) ;  re- 
peat the  same  construction  for  two  other  vertices  B,  b;  one  of  the 
points  in  which  the  two  circumferences  intersect  will  be  the  point 
sought. 

CENTERS  OF  SIMILITUDE  IN  CIRCLES. 

THEOREM. 

Two  circles,  as  well  as  two  regular  polygons  of  an  even  number  of 
sides,  have  two  centers  of  similitude,  the  one  internal  and  the  other  ex- 
ternal. 

When  the  two  circles  are  exterior  to  each  other,  prove  that  the 
points  in  which  their  common  tangents  meet  are  centers  of  simili- 
tude. This  point  may  be  found  by  dividing  the  line,  joining  the  cen- 
ters in  the  ratio  of  the  radii. 

When  the  circles  touch  each  other  externally,  prove  that  the  point  of 
contact  is  an  internal  center  of  similitude  ;  and  that  if  they  touch  each 
other  internally,  the  point  of  contact  is  an  external  center  of  simil- 
itude. 

There  exist  several  other  remarkable  particular  cases. 

For  two  concentric  circles,  the  centers  of  similitude  unite  in  the 
common  center. 

For  two  equal  circles,  the  internal  center  of  similitude  is  at  the 


APPENDIX    II.  O 

middle  of  the  line  joining  their  centers  ;  the  external  is  at  an  infinite 
distance. 

When  one  of  the  circles  degenerates  into  a  right  line,  1°.  The  cen- 
ters of  similitude  are,  at  die  i -xtremities  of  a  diameter  of  the  other 
circle,  perpendicular  to  the  line.  2°.  If  one  of  the  circles  reduces 
to  a  point,  that  point  is  itself  the  center  of  similitude,  both  internal 
and  external. 

Scholium.  To  be  proved.     When  three  circles  are  situated  upon 

the  same  plane  which  gives  six  centers  of  similitude  ;  1°.   The  three 

d  centers  of  similitude  ;  2°.   One  external  and  two  internal — are 

vpon  a  same  line,  which  gives  four  lines,  passing  through  six  point3 

combined,  three  and  three. 

RADICAL  AXIS  AND  RADICAL  CENTER. 

Definitions.  A  radical  axis  of  two  circles  is  the  locus  of  points  from 
each  of  which  equal  tangents  can  be  drawn  to  the  two  circles. 

Construction.  Divide  the  line  joining  the  centers  of  the  two  circles 
in  such  a  manner  that  the  difference  of  the  squares  of  the  two  parts 
is  equal  to  the  difference  of  the  squares  of  the  radii,  and  the  perpen- 
dicular to  this  line  at  the  point  of  division  will  be  a  radical  axis. 

Trove  that  to  find  the  point  on  the  line  joining  the  centers  it  is 
only  necessary  to  lay  off  from  the  middle  of  this  line,  on  the  side 
toward  the  smaller  circle,  a  distance  equal  to  half  a  third  propor- 
tional to  the  distance  between  the  centers  and  the  square  root  of  the 
difference  of  the  squares  of  the  radii.     (See  Prob.  14,  p.  80.) 

Each  particular  case,  however,  presents  a  more  simple  construction. 
1°.  If  the  circles  be  exterior,  or  in  any  position  for  which  there  exists 
I  common  tangent,  as  the  middle  point  of  the  portion  of  this  tangent 
comprehended  between  the  two  points  of  contact  belongs  to  the  rad- 
ical axis,  we  draw  through  this  point  a  perpendicular  to  the  line  join- 
iii''  the  centers  of  the  circles,  and  thus  have  this  axis.  2°.  When  the 
circles  touch  either  exteriorly  or  interiorly,  the  common  point  of  the 
two  circumferences  belongs  necessarily  to  the  radical  axis,  and  thus 
leads  to  its  determination,  as  before.  It  is  then  the  common  tangent 
to  the  circles  at  this  point.  3°.  When  the  circles  cut  each  other,  the 
common  chord  produced  both  ways  is  the  radical  axis. 

Concentric  circles  have  no  radical  axis.  When  the  two  circles  are 
equal,  the  radical  axis  is  the  perpendicular  at  the  middle  of  the  line 
joining  the  centers. 

h'  one  of  the  circles  be  reduced  to  a  point,  the  radical  axis  is  ob- 
tained by  joining  the  middles  of  the  tangents  drawn  from  this  point  to 
the  other  circle. 


6  GEOMETRY. 

If  one  of  the  circles  degenerates  into  a  right  line,  the  radical  axis  is 
the  line  itself. 

Radical  Center. —  Three  circles  situated  in  the  same  plane  (the 
centers  of  which  are  not  in  the  same  line)  give,  by  their  combination 
two  and  two,  three  radical  axes ;  and  these  three  axes  cut  each  other 
in  the  same  point. 

For  the  two  first  cutting  each  other,  and  being  respectively  perpen- 
dicular to  two  lines  which  cut,  their  point  of  intersection  is  such  that 
there  can  be  drawn  from  this  point  to  the  three  circumferences  equal 
tangents ;  consequently,  it  belongs  to  the  third  radical  axis. 

This  point  is  called  the  radical  center  of  the  three  circles. 

From  this  definition,  and  from  what  has  been  shown  above,  it  fol- 
lows, that  if  three  circumferences  intersect,  the  three  chords  which 
unite  their  points  of  intersection  meet  in  the  same  point. 

When  this  point  of  intersection  is  exterior  to  the  three  circles,  the 
six  tangents  from  this  point  are  equal. 

Theorem.  Prove  that  if,  from  any  point  of  a  radical  axis  of  two  cir- 
cles, a  secant  be  drawn  meeting  the  circumferences  in  four  points,  these 
four  points  will  be  in  the  circumference  of  a  third  circle. 

This  may  be  proved  by  aid  of  the  theorem  that  the  rectangle  of  a 
secant  and  its  external  segment  is  constant  (th.  42),  together  with 
the  construction  of  a  radical  axis. 

Theorem.  If  through  one  of  two  centers  of  similitude  (external  or 
internal)  of  two  circles  two  secants  to  these  circles  be  drawn,  1°.  The 
eight  points  of  intersection  combined,  four  and  four,  in  a  suitable  manner, 
form  four  groups,  situated  respectively  upon  as  many  new  circumferen- 
ces; 2°.  These  four  circumferences  have  for  a  common  radical  center 
that  center  of  similitude  which  served  to  determine  these  circumferences. 

Note.  The  above  theory  will  be  found  of  great  use  in  the  solution 
of  all  problems  involving  the  contact  of  circles. 

CONJUGATE  POINTS,  POLES,  AND  POLAR  LINES. 

Conjugate  points  are  two  points  situated  the  one  within,  the  other 
without,  a  circle,  in  such  a  manner  that  the  distances  of  every  point  in 
the  circumference  from  these  two  points  are  in  a  constant  ratio.  The 
circle  is  called  the  regulating  circle. 

The  point  within  the  circle  being  given,  to  determine  its  conjugate, 
erect  at  the  given  point  a  perpendicular  to  the  line  joining  the  given 
point  and  the  center,  and  at  the  point  where  this  perpendicular  meets 
the  circumference  draw  a  tangent  which  will  meet  the  line  joining 
the  given  point  and  center  produced  in  the  point  required.  Prove 
this. 


APPENDIX   II.  7 

A  chord  of  contact  is  a  line  joining  the  points  of  contact  of  two 
nts  drawn  from  the  same  point. 

Theorkm.  The  chords  of  contact  of  all  tangents  which  meet  in  one 
and  the  same  line  will  meet  in  the  same  point,  and  the  conjugate  of 
this  point  is  the  loot  of  a  perpendicular  let  fall  from  the  center  of  the 
circle  upon  the  line  in  which  the  tangents  meet. 

Tin*  point  in  which  all  the  chords  of  contact  meet  is  called  a  pole, 
and  the  line  in  which  the  tangents  all  meet,  a  polar  line. 


j 


GEOMETRY  OF  PLANES* 


DEFINITIONS. 

1.  The  angle  formed  by  two  lines  not  in  the  same 
plane  is  the  angle  formed  by  one  of  them  with  a  line 
drawn  through  any  point  of  it  parallel  to  the  other. 

2.  A  plane  is  a  surface  in  which,  if  any  two  points 
be  taken,  the  straight  line  which  joins  these  points 
will  be  wholly  in  that  surface. 

3.  A  straight  line  is  said  to  be  perpendicular  to  a 
plane  when  it  is  perpendicular  to  all  the  straight  lines 
in  the  plane  which  pass  through  the  point  in  which  it 
meets  the  plane. 

This  point  is  called  the  foot  of  the  perpendicular. 

4.  The  inclination  of  a  straight  line  to  a  plane  is 
the  acute  angle  contained  by  the  straight  line,  and 
another  straight  line  drawn  from  the  point  in  which 
the  first  meets  the  plane,  to  the  point  in  which  a  per- 
pendicular to  the  plane,  drawn  from  any  point  in  the 
first  line,  meets  the  plane. 

5.  A  straight  line  is  said  to  be  parallel  to  a  plane 
when  it  can  not  meet  the  plane,  to  whatever  distance 
both  be  produced. 

6.  It  will  be  proved  in  Prop.  2,  that  the  common 
intersection  of  two  planes  is  a  straight  line ;  this  be- 
ing premised, 

The  angle  contained  by  two  planes,  which  cut  one 
another,  is  measured  by  the  angle  contained  by  two 
jhtraight  lines  drawn,  one  in  each  of  the  planes,  per- 
pendicular to  their  common  intersection  at  the  same 
^point. 

This  angle  may  be  acute,  right,  or  obtuse. 

*  Students  intending  to  pursue  that  subject,  may  here  with  advan- 
tage take  up  Plane  Trigonometry  before  going  on  with  the  Geome- 
try of  Planes  and  Solids. 


a  GEOMETRY. 

If  it  be  a  right  angle,  the  planes  are  said  to  be  per- 
pendicular to  each  other. 

The  angle  formed  by  two  planes  is  called  diedral. 

7.  Two  planes  are  parallel  to  each  other  when 
they  can  not  meet,  to  whatever  distance  both  be  pro- 
duced. 

8.  A  plane  is  ordinarily  represented,  in  a  diagram, 
by  a  parallelogram,  and  called  by  the  two  letters  at 
the  opposite  (diagonally)  angle.  This  plane,  which 
must  be  conceived  to  be  indefinitely  extended,  divides 
space  into  two  indefinite  portions  called  regions. 

Two  or  more  planes  are  represented  in  their  rela- 
tive position  not  accurately,  but  by  a  sort  of  perspec- 
tive. 

PROP.    I. 

A  straight  line  can  not  be  partly  in  a  plane  and  part- 
ly out  of  it. 

For,  by  def.  (1),  when  a  straight  line  has  two  points 
common  with  a  plane,  it  lies  wholly  in  that  plane. 


D 


PROP.    II. 

If  two  planes  cut  each  other,  their  common  intersec- 
tion is  a  straight  line. 

Let  the  two  planes  AB,  CD  cut  each 
other,  and  let  P,  Q  be  two  points  in 
their  common  section. 

Join  P,  Q ; 

Then,  since  the  points  P,  Q  are  in 
the  same  plane  AB,  the  straight  line 
PQ  which  joins  them  must  lie  wholly 
in  that  plane  (def.  2). 

For  a  similar  reason,  PQ,  must  lie 
wholly  in  the  plane  CD. 

.*.  The   straight  line   PQ  is  common   to  the  two 
planes,  and  is  .*.  their  common  intersection. 

Note.  In  this  and  the  following  diagrams  concealed 
lines  are  drawn  dotted. 


GEOMETRY    OF    PLANES. 


PROP.    III. 


Any  number  of  planes  may  be  drawn  through  the 
same  straight  line. 


■.-. 


For  let  a  plane,  drawn  through  a  straight  line,  be 
conceived  to  revolve  round  the  straight  line  as  an 
axis.  Then  the  different  positions  assumed  by  the 
revolving  plane  will  be  those  of  different  planes 
drawn  through  the  straight  line 

PROP.    IV. 

One  plane,  and  one  plane  only,  can  be  drawn, 

1°.  Through  a  straight  line,  and  a  point  not  sit- 
uated in  the  given  line. 

2°.  Through  three  points  which  are  not  in  the 
same  straight  line. 

3°.  Through  two  straight  lines  which  intersect 
each  other. 

4°.   Through  two  parallel  straight  lines. 

1.  For  if  a  plane  be  drawn  through  the  given  line, 
and  be  conceived  to  revolve  round  it  as  an  axis,  it 
must  in  the  course  of  a  complete  revolution  pass 
through  the  given  point,  and  so  assume  the  position 
enounced  in  1°. 

Also,  one  plane  only  can  answer  these  conditions, 
for  if  we  suppose  a  second  plane  passing  through  the 
same  straight  line  and  point,  it  must  have  at  least  two 
intersections  with  the  first,  which  is  evidently  impos- 
sible. 

2.  Join  two  of  the  points  ;  this  case  is  then  reduced 
to  the  last. 

.'}.  Take  a  point  in  each  of  the  lines  which  is  not 
the  point  of  intersection  ;  join  these  two  points  ;  the 
case  is  now  the  same  as  the  two  former. 

4.  Parallel  straight  lines  are  in  the  same  plane,  and, 
by  the  first  case,  one  plane  only  can  be  drawn  through 
either  of  them,  and  a  point  assumed  in  the  other. 


GEOMETRY. 

Cor.  Hence  the  position  of  a  plane  is  determined  by, 

1.  A  straight  line,  and  a  point  not  in  the  given 

straight  line. 

2.  A  triangle,  or  three  points  not  in   the  same 

straight  line. 

3.  Two  straight  lines  which  intersect  each  other. 

4.  Two  parallel  straight  lines. 


prop.  v. 


If  a  straight  line  be  perpendicular  to  two  other 
straight  lines  which  intersect  at  its  foot  in  a  plane,  it 
will  be  perpendicular  to  every  other  straight  line  drawn 
through  its  foot  in  the  same  plane,  and  will  therefore 
be  perpendicular  to  the  plane. 

Let  XZ  be  a  plane,  and  let  the 
straight  line  PQ,  be  perpendicular 
to  the  two  straight  lines  AB,  CD 
which  intersect  in  Q  in  the  plane 
XZ. 

We  shall  prove  that  PQ,  will  be 
perpendicular  to  any  other  straight 
line  EF,  drawn  through  Q  in  the 
plane  XZ. 

Draw  through  any  point  K  in  QE  a  straight  line 
GH,  such  that  GK  =  KH.     (See  exercise  4,  p.  72.) 

Join  P,  G  ;  P,  K  ;  P,  H  ; 

Then,  since  GH,  the  base  of  the  A  GQH,  is  bisect- 
ed in  K  ; 
.-.  (by  th.  30),  GQ2  +  HQ2  =  2GK2  +  2QK2  ...  (1) 

Similarly,  since  GH,  the  base  of  A  GPH,  is  bisect- 
ed in  K ; 

...  Gp  +  HF  =  2GKa  +  2pK9 

But  the  triangles  PQG,  PQH  are  right-angled  at 
Q;  .*.  the  last  expression  becomes 
PQ2  +  GQ2  +  PQ2  +  HQ2  =  2GK2  +  2PK3  ...  (2) 
Taking  (1)  from  (2),  there  remains 
2PQ2  =  2PK2  —  2QK2, 
.*.  dividing  by  2,  and  transposing, 
PQ2  +  QK2  =  PK2. 


GEOMETRY    OF    PLANES. 


Henco  the  triangle  PQK  is  right-angled  at  Q,  for  in 
the  right-angled  triangle  alone  the  sum  of  the  squares 
pf  two  <>f  the  sides  is  equal  to  the  square  of  the  third 
theorems  26,  28,  2i>.) 
In  like  manner,  it  may  be  proved  that  PQ  is  at 
right  angles  to  every  other  straight  line  passing 
through  Q  in  the  plane  XZ. 


PROP.  VI. 


A  perpendicular  is  the  shortest  line  which  can  be 
drawn  to  a  plane  from  a  point  without. 

Let  PQ  be  perpendicular  to  the  p 

plane  XZ  ; 

From  P  draw  any  other  straight 
line  PK  to  the  plane  XZ  ; 
Then  PQ  <  PK. 

In  the  plane  XZ  draw  the  straight 
line  QK,  joining  the  points  Q,  K. 

Then,  since  the  line  PQ  is  perpen- 
dicular to  the  plane  XZ,  it  is  per- 
pendicular to  QK,  a  line  of  the  plane  ;  and  .*.  PQ  is 
less  than  PK.     (Geom.  Theor.,  17.) 


frop.  vir. 

Oblique  lines  equally  distant  from  the  perpendicular 
are  equal,  and,  if  two  oblique  lines  be  unequally  distant 
from  the  perpendicular,  the  more  distant  is  the  larger. 

That  is,  ifQG,  Qll,  QK r 

are  all  equal,  then  PG,  PH,  PK  .  .  . 
.  .  .  are  all  equal ;  and  if  QI  be  great- 
er than  QG,  then  PI  is  greater  than 
PG.  For  the  three  right-angled  tri- 
angles PQG,  PQH,  PQK  baying  two 
sides  in  each  equal,  the  third  sides 
are  equal  (th.  2G,  corol.  2)  ;  and 
since  PH  <  PI  (th.  17), .-.  PG  <  PL* 

*  Th  tion  pfibrdi  a  method  of  finding  the  foot  of  a  per- 

il:u-  t<»  ;i  ffiveu  plaae  from  a  point  without.     With  a  straight 


6 


GEOMETKY. 


Cor.  A  perpendicular  measures  the  distance  of  any 
point  from  a  plane.  The  distance  of  one  point  from 
another  is  measured  by  the  straight  line  joining  them, 
because  this  is  the  shortest  line  which  can  be  drawn 
from  one  point  to  another.  So,  also,  the  distance  from 
a  point  to  a  line  is  measured  by  a  perpendicular,  be- 
cause this  line  is  the  shortest  that  can  be  drawn  from 
the  point  to  the  line.  In  like  manner,  the  distance 
from  a  point  to  a  plane  must  be  measured  by  a  per- 
pendicular drawn  from  that  point  to  the  plane,  be- 
cause this  is  the  shortest  line  that  can  be  drawn  from 
the  point  to  the  plane. 


PROP.    VIII. 


If,  from  a  given  point  without  a  plane,  a  perpendic- 
ular be  let  fall  to  the  plane,  and  from  its  foot  a  perpen- 
dicular be  drawn  to  a  line  of  the  plane,  and  the  point 
of  intersection  be  joined  with  the  point  without,  the  last 
line  will  be  perpendicular  to  the  line  of  the  plane. 

Let  PQ  be  a  perpendicular  on  the  plane  XZ,  and 
GH  a  straight  line  in  that  plane ;  if  from  Q,  the  foot 
of  the  perpendicular,  QK  be  drawn  perpendicular  to 
GH,  and  P,  K  be  joined  ;  then  PK  will  be  perpendic- 
ular to  GH. 

Take  KG  =  KH ;  join  P,  G ;  P,  p 

H  ;  Q,  G  ;  Q,  H  ; 

Because  KG  =  KH,  and  KQ  is 
common  to  the  triangles  GQK, 
HQK,  and  the  angle  GKQ  =  an- 
gle HKQ,  each  being  a  right  an- 
gle* 

.-.  QG  =  QH, 

.\  PG  ~  PH  (last  Prop.),  z 

Hence  the  two  triangles  GKP,  HKP  have  the  two 
sides  GK,  KP  equal  to  the  two  sides  HK,  KP,  and 

rod,  one  end  of  which  is  fixed  at  the  given  point,  touch  the  given 
plane  in  three  points  not  in  the  same  right  line  ;  find  the  center  of  the 
circle  passing  through  these  three  points,  and  it  will  be  the  foot  of 
the  perpendicular  required. 


GEOMETRY    OF    PLANES.  7 

the  remaining  side  GP,  equal  to  the  remaining  side 

HP. 

.-.  Angle  GKP  =  angle  HKP,  and  .*.  each  of  them 
is  a  right  angle  (def.  12). 

Cor.  GH  is  perpendicular  to  the  plane  PQK,  for 
GH  is  perpendicular  to  each  of  the  two  straight  lines 
Kl\  KQ  (Prop.  5). 

Remark. — The  two  straight  lines  PQ,  GH  present 
an  example  of  two  straight  lines  which  do  not  meet, 
because  they  are  not  situated  in  the  same  plane. 

The  shortest  distance  between  these  two  lines  is 
the  straight  line  QK,  which  is  perpendicular  to  each 
of  them. 

For,  join  any  two  other  points,  as  P,  G ; 
Then  PG  >  PK  )  ,    f  ^ 

And  KP>KQilastPr°P' 

PG  >  KQ. 

The  two  lines  PQ,  GH,  although  not  situated  in 
the  same  plane,  are  considered  to  form  a  right  angle 
with  each  other.  For  PQ,  and  a  straight  line  drawn 
through  any  point  in  PQ  parallel  to  GH,  would  form 
a  right  angle. 

In  like  manner,  PG  and  QK,  which  represent  any 
two  straight  lines  not  situated  in  the  same  plane,  are 
considered  to  form  with  each  other  the  same  angle 
which  PG  would  make  with  any  parallel  to  QK, 
drawn  through  a  point  in  PG. 

PROP.  IX. 

If  two  straight  lines  be  perpendicular  to  the  same 
plane,  they  will  be  parallel  to  each  other. 

Let  each  of  the  straight  lines  PQ, 
GH  be  perpendicular  to  the  plane  XZ. 

Then  PQ  will  be  parallel  to  GH. 

In  the  plane  XZ  draw  the  straight 
line  QH,  joining  the  points  Q,  H. 

Then,  since  PQ,  GH  are  perpen- 
dicular to  the  plane  XZ,  they  are  per- 
pendicular to  the  straight  line  QH  in 


8  GEOMETRY. 


that  plane ;  and,  since  PQ,  GH  are  both  perpendicu- 
lar to  the  same  line  QH,  they  have  the  same  direc- 
tion and  are  parallel  to  each  other. 


prop.  x. 


Conversely,  if  two  straight  lines  be  parallel,  and  if 
one  of  them  be  perpendicular  to  any  plane,  the  other 
will  also  be  perpendicular  to  the  same  plane. 

For  let  GH,  PQ  be  the  two  lines.  Draw  through 
P  a  perpendicular  to  the  plane  ;  this,  by  the  last  Prop., 
will  be  parallel  to  GH,  and  must,  therefore,  be  identi- 
cal with  QP,  since  through  a  given  point  but  one 
parallel  can  be  drawn  to  a  given  line. 

Cor.  Two  straight  lines  parallel  to  a  third  are  par- 
allel to  each  other. 

For,  conceive  a  plane  perpendicular  to  any  one  of 
them,  then  the  other  two  being  parallel  to  the  first, 
will  be  perpendicular  to  the  same  plane  ;  hence,  by 
the  last  Prop.,  they  will  be  parallel  to  each  other. 

The  three  straight  lines  are  not  supposed  to  be  in 
the  same  plane. 

This  corollary  follows,  also,  from  our  definition  of 
parallel  lines  (def.  8,  p.  1). 

PROP.  XI. 

If  a  straight  line,  without  a  given  plane,  be  parallel 
to  a  straight  line  in  the  plane,  it  will  be  parallel  to  the 
plane. 

Let  AB,  lying  without  the  plane        a  b 

XZ,  be  parallel  to  CD,  lying  in  the 
plane.  I 

Then  AB  is  parallel  to  the  plane 

xz.  rtrhx 

Through   the   parallels  AB,  CD    1     e  g    \ 

pass  the  plane  ABCD,  and  suppose    L \ 

it,  as  well  as  the  lines  AB  and  CD,    z 
to  extend  indefinitely. 

If  the  line  AB  can  meet  the  plane  XZ,  it  must  meet 


GEOMETRY    OK    FLAXES.  9 

it  in  some  point  of  the  line  CD,  which  is  the  common 
intersection  of  the  two  planes,  for  the  line  AB  can 
not  get  out  of  the  plane  AD  (def.  2). 

But  AB  can  not  meet  CD,  because  AB  is  parallel 
to  CD. 

Hence  AB  can  not  meet  the  plane  XZ,  i.  e.,  AB  is 
parallel  to  the  plane  XZ  (def.  5). 

PROP.   XII. 

The  sections  made  by  a  plane  cutting  two  parallel 
planes  are  parallel. 

Let   FE,   GH    be   the   sections 
made  by  the  plane  GF  which  cuts    r 
the  parallel  planes  XZ,  WY ;  L  . 

Then  FE  will  be  parallel  to  GH.    a 

For  if  the  lines  FE,  GH,  which 
are  situated  in  the  same  plane,  be  w 
not  parallel,  they  will  meet  if  pro-    r- 
duced.     Therefore,  the  planes  XZ,    \ 
WY,  in  which  these  lines  lie,  will 
meet  if  produced,  and  .-.  can  not  be  parallel,  which 
is  contrary  to  the  hypothesis. 

.-.  FE  is  parallel  to  GH. 

PROP.   XIII. 

Parallel  straight  lines  included  between  two  parallel 
planes  are  equal. 

Let  (see  last  fig.)  the  parallels  EG,  FH  be  cut  by 
the  parallel  planes  XZ,  WY,  in  the  points  G,  H,  E,  F. 

Then  EG  =--  FH, 

Through  the  parallels  EG,  FH,  draw  the  plane 
EFGH,  intersecting  the  parallel  planes  in  GH,  FE. 

Then    GH  is  parallel  to  FE,  by  last  Prop. 

And      GE  is  parallel  to  HF  (by  hyp.) ; 
.*.  GHFE  is  a  parallelogram  ;  and,  therefore, 
EG  =  FII. 

Cor.  Two  parallel  planes  are  every  where  equidis- 
tant. For  the  perpendiculars  which  measure  their 
distance,  being  parallels,  are  every  where  equal. 


10  GEOMETRY. 


prop.  XIV. 


X 

w 

r 

Ai — 7 

if- 

i 

IB 

z 

*Y 

If  two  planes  be  parallel  to  each  other,  a  straight 
line  which  is  perpendicular  to  one  of  the  planes  will  be 
perpendicular  to  the  other  also. 

Let  the  two  planes  XZ,  WY  be 
parallel,  and  let  the  straight  line  AB 
be  perpendicular  to  the  plane  XZ  ; 

Then  will  AB  be  perpendicular 
to  WY. 

For,  from   any   point  H   in  the 
plane  WY,  draw  HG  perpendicular  jf 
to  the  plane  XZ,  and  draw  AG,  BH. 

Then,  since  BA,  HG  are  both  perpendicular  to  XZ, 
they  are  (Prop.  9)  parallel  to  each  other. 

And,  since  the  planes  XZ,  WY  are  parallel,  the 
parallels  BA,  HG  are  (by  the  last  Prop.)  equal. 

'Hence  (th.  21)  AG  is  parallel  to  BH  ;  and  AB,  be- 
ing perpendicular  to  AG,  is  perpendicular  to  its  par- 
allel BH  also. 

In  like  manner,  it  may  be  proved  that  AB  is  per- 
pendicular to  any  other  line  which  can  be  drawn 
from  B  in  the  plane  WY. 

.-.  AB  is  perpendicular  to  the  plane  WY. 

PROP.   xv. 

Conversely,  if  two  planes  be  perpendicular  to  the 
same  straight  line,  they  will  be  parallel  to  each  other. 

For  if  they  could  meet,  from  any  common  point 
of  the  two  planes  draw  two  lines,  one  in  each  plane, 
to  the  extremities  of  the  line  to  which  they  are  both 
perpendicular  ;  we  should  thus  have  two  perpendic- 
ulars from  the  same  point  to  the  same  line,  which  is 
impossible. 

PROP.  XVI. 

If  two  straight  lines  which  form  an  angle  be  parallel 
to  two  other  straight  lines  which  form  an  angle  in  the 


9EOMETEY  OF   PLANES.  11 

same  direction,  although  not  in  the  same  plane  with  the 
former,  the  tiro  angles  will  be  equal,  and  their  planes 
will  be  parallel. 

Let  the  two  straight  lines  AB,  n 

BC,  in  the  plane  XZ,  be  parallel  j 

to  the  two  DE,  EF,  in  the  plane     x 

Then  angle  ABC  -  angle  DEF.   / !*«-  ±^a_A 

For,  make  BA  =  ED,  BC  =  EF ;  z 
join  A,  C ;  D,  F;  A,  D  ;  B,  E;   w 
C,  F;  i— 

Then  the  straight  lines  AD,  BE,         » 

which  join  the  equal  and  parallel    ' ^^ — v 

straight  lines  AB,  DE,  are  them- 
selves equal  and  parallel. 

For  the  same  reason,  CF,  BE  are  equal  and  paral- 
lel. 

.-.  AD,  CF  are  equal  and  parallel  (cor.,  Prop.  10.), 
and  .\  AC,  DF  are  also  equal  and  parallel  (th.  21). 

Hence  the  two  triangles  ABC,  DEF,  having  ali 
their  sides  equal,  each  to  each,  have  their  angles  also 
equal. 

.-.  angle  ABC  =  angle  DEF. 

Again,  the  plane  XZ  is  parallel  to  the  plane  WY. 

For,  if  not,  let  a  plane  drawn  through  A,  parallel 
to  DEF,  meet  the  straight  lines  FC,  EB  in  G  and  H. 

Then  D A  =  EH  =  FG  (Prop.  1 3). 

But  DA  =  EB=FC 

.-.EH  =  EB,     FG  =  FC, 
which  is  absurd. 

Cor.  1.  If  two  parallel  planes  XZ,  WY  are  met  by 
two  other  planes  ADEB,  CFEB,  the  angles  ABC, 
DEF,  formed  by  the  intersection  of  the  parallel  planes, 
will  be  equal. 

For  the  section  AB  is  parallel  to  the  section  DE 
(Prop.  12). 

So,  also,  the  section  BC  is  parallel  to  the  section  EF. 
.-.  angle  ABC  =  angle  DEF. 

Cor.  2.  If  three  straight  lines  AD.  BE,  CF,  not  sit- 


12 


GEOMETRY. 


X       c 

r-?t  \ 

-J  / 1  z 

P    4 1 

uated  in  the  same  plane,  be  equal  and  parallel,  the  tri- 
angles ABC,  DEG,  formed  by  joining  the  extremities 
of  these  straight  lines,  will  be  equal,  and  their  planes 
will  be  parallel. 

mor.  xvu. 

If  two  straight  lines  be  cut  by  parallel  planes,  they 
will  be  cut  in  the  same  ratio. 

Let  the  straight  lines  AB,  CD  be 
cut  by  the  parallel  planes  XZ,  WY, 
VS,  in  the  points  A,  E,  B  ;  C,  F,  D  ; 

Then     AE:EB::CF:FD. 

Join  A,  C ;  B,  D  ;  A,  D  ;  and  let 
AD  meet  the  plane  WY  in  G  ;  join 
E,  G  ;  G,  F ; 

Then  the  intersections  EG,  BD 
of  the  parallel  planes  WY,  VS, 
with  the  plane  ED,  are  parallel 
(Prop.  12). 

.\  AE  :  EB  : :  AG  :  GD  (th.  Gl). 

Again   the   intersection  AC,   GF  of  the   parallel 
planes  XZ,  YW,  with  the  plane  CG,  are  parallel. 
.-.  AG:GD::CF:FD. 

.*.  substituting  the  second  ratio  for  the  first  of  this 
proportion  in  the  previous  proportion,  we  have 
AE  :  EB  : :  CF :  FD. 

PROP.   XVIII. 

If  a  straight  line  be  perpendicular  to  a  plane,  every 
plane  which  passes  through  it  will  be  at  right  angles  to 
that  plane. 

Let  the  straight  line  PQ  be  perpen- 
dicular to  the  plane  XZ. 

Through  PQ,  draw  any  plane  PO, 
intersecting  XZ  in  the  line  OQW. 

Then  the  plane  PO  is  perpendicu- 
lar to  the  plane  XZ. 

For,  draw  RS,  in  the  plane  XZ, 
perpendicular  to  WQO. 

Then,  since  the  straight  line  PQ  is 


ffr 

\ 

X 

\° 

GEOMETRY    OF    PLANES. 


13 


perpendicular  to  the  plane  XZ,  it  is  perpendicular  to 
the  two  straight  lines  RS,  OW,  which  pass  through 
its  foot  in  that  plane. 

But  the  angle  PQR  is  contained  between  PQ,  QR, 
which  are  perpendiculars  at  the  same  point  to  OW, 
the  common  intersection  of  the  planes  XZ,  PO ;  this 
angle,  therefore,  measures  the  angle  of  the  two  planes 
(def.  6) ;  hence,  since  this  angle  is  a  right  angle,  the 
two  planes  are  perpendicular  to  each  other. 

Cor.  If  three  straight  lines,  such  as  PQ,  RS,  OVV, 
be  perpendicular  to  each  other,  each  will  be  perpen- 
dicular to  the  plane  of  the  other  two,  and  the  three 
planes  will  be  perpendicular  to  one  another. 


PROP.  XIX. 

If  two  planes  be  perpendicular  to  each  other,  a 
Straight  line  drawn  in  one  of  the  planes  perpendicular 
to  their  common  section  will  be  perpendicular  to  the 
other  plane. 

Let  the  plane  VO  be  perpendicular 
to  the  plane  XZ,  and  let  OW  be  their 
common  section. 

In  the  plane  VO  draw  PQ  perpen- 
dicular to  OW ; 

Then  PQ  is  perpendicular  to  the 
plane  XZ. 

For,  from  the  point  Q,  draw  QR 
in  the  plane  XZ,  perpendicular  to 
OW. 

Then,  since  the  two  planes  are  perpendicular,  the 
angle  PQR  is  a  right  angle  (def.  f>). 

.*.  The  straight  line   PQ  is  perpendicular  to  the 
straight  lines  QR,  QO,  which  intersect  at  its  foot  in 
the  plane  XZ. 
.*.  PQ  is  perpendicular  to  the  plane  XZ  (Prop.  5). 

Cor.  If  the  plane  VO  be  perpendicular  to  the  plane 
XZ,  and  if  from  any  point  in  OW,  their  common  in- 
tersection, we  erect  a  perpendicular  to  the  plane  XZ, 
that  straight  line  will  lie  in  the  plane  VO. 


X 

14 


GEOMETRY. 


For  if  not,  then  we  may  draw  from  the  same  point 
a  straight  line  in  the  plane  VO,  perpendicular  to  O  W, 
and  this  line,  by  the  Prop.,  will  be  perpendicular  to 
the  plane  XZ. 

Thus  we  should  have  two  straight  lines  drawn  from 
the  same  point  in  the  plane  XZ,  each  of  them  perpen- 
dicular to  this  plane,  which  is  impossible. 


PROP.  xx. 


If  tivo  planes  which  cut  each  other  be  each  of  them 
perpendicular  to  a  third  plane,  their  common  section 
will  be  perpendicular  to  the  same  plane. 

Let  the  two  planes  VO,  TW,  whose 
common  section  is  PQ,  be  both  per- 
pendicular to  the  plane  XZ. 

Then  PQ,  is  perpendicular  to  the 
plane  XZ. 

For,  from  the  point  Q,  erect  a  per- 
pendicular to  the  plane  XZ. 

Then,  by  cor.  to  last  Prop.,  this 
straight  line  must  be  situated  at  once 
in  the  planes  VO  and  TW,  and  is  . 
section. 


w 


\ 


their 


common 


EXERCISES. 


1.  Prove  that  but  one  plane  can  be  passed  through  a  given  point 
perpendicular  to  a  given  line. 

2.  Prove  that  but  one  perpendicular  can  be  drawn»from  a  given 
point  to  a  given  plane. 

3.  That  when  a  plane  is  perpendicular  at  the  middle  of  a  given 
line,  every  point  of  the  plane  is  equally  distant  from  the  extremities 
of  the  line,  and  that  every  point  out  of  the  plane  is  unequally  distant. 

4.  That  through  a  given  line  in  a  plane  only  one  plane  perpendic- 
ular to  the  given  plane  can  be  passed. 

5.  That  through  a  line  parallel  to  a  given  plane  but  one  plane  can 
be  passed  perpendicular  to  the  given  plane. 

6.  That  if  two  planes  which  intersect  contain  two  lines  parallel  to 
each  other,  the  intersection  of  the  planes  will  be  parallel  to  the  lines. 

7.  That  if  a  line  be  parallel  to  a  plane,  every  other  plane  passed 


EXERCISES.  15 

through  this  line  and  meeting  the  former,  will  intersect  it  hi  a  second 
line  parallel  to  the  first. 

8.  That  when  a  line  is  parallel  to  one  plane  and  perpendicular  to 
another,  the  two  planes  are  perpendicular  to  each  other. 

9.  That  a  line  parallel  to  a  plane  is  every  where  equally  distant 
from  that  plane.     The  same  of  two  parallel  planes. 

10.  That  two  lines  are  always  either,  in  one  and  the  same  plane  or 
two  parallel  planes. 

Note. — These  planes,  the  system  of  which  is  unique  for  each  system 
of  two  lines  not  situated  in  the  same  plane,  are  called  the  parallel 
planes  of  these  lines. 

11.  Show  that  but  one  plane  can  be  drawn  through  a  given  point 
parallel  to  a  given  plane. 

12.  l'rove  that  two  plane's  parallel  to  a  third  are  parallel  to  each 
other. 

13.  Draw  a  perpendicular  to  two  lines  not  in  the  same  plane. 

14.  Prove  that  if  two  lines  are  parallel  in  space,  and  planes  be 
passed  through  them  perpendicular  to  a  third  plane,  the  two  planes 
will  be  parallel. 

15.  That  if  a  line  be  parallel  to  one  of  two  perpendicular  planes, 
and  a  plane  be  passed  through  the  line  perpendicular  to  the  other 
plane,  it  will  be  parallel  to  the  first  plane. 

16.  To  place  a  perpendicular  to  a  given  plane  at  a  given  point  of 
the  plane. 

17.  To  place  a  plane  perpendicular  to  a  given  plane,  and  intersect- 
ing it  in  a  given  line. 

18.  To  place  a  plane  parallel  to  a  given  plane. 

19.  To  place  a  line  under  a  given  angle  to  a  given  plane. 

20.  To  place  a  plane  under  a  given  angle  to  a  given  plane,  and  in- 
tersecting it  hv-a  given  line. 

21.  To  pk^fcfa  plane  perpendicular  to  two  give  planes. 


POLYHEDRAL  ANGLES. 


DEFINITION. 

A  polyhedral  angle,  improperly  called  a  solid  angle, 
is  the  angular  space  contained  between  several  planes 
which  meet  in  the  same  point.  This  point  is  called 
the  vertex. 

Three  planes  at  least  are  required  to  form  a  poly- 
hedral angle. 

A  polyhedral  angle  is  called  a  trihedral,  tetrahedral, 
&c,  angle,  according  as  it  is  formed  by  three,  four, 
&c,  plane  angles. 

A  polyhedral  angle  is  named  from  the  letter  at  its 
vertex. 

A  polyhedral  angle  is  called  regular  when  all  its  plane  angles  are 
equal  and  all  its  diedral  angles  equ;il. 

A  trihedral  is  called  birectangular,  ti-irectangular,  when  two  or 
three  of  its  diedral  angles  are  right  angles. 

When  two  of  the  diedral  angles  are  equal,  it  is  called  isohedral. 

PROP.   I. 

If  a  polyhedral  angle  be  contained  by  three  plane 
angles,  the  sum  of  any  two  of  these  angles  will  be  great- 
er than  the  third. 

It  is  unnecessary  to  demonstrate  this  proposition, 
except  in  the  case  where  the  plane  angle,  which  is 
compared  with  the  two  others,  is  greater  than  either 
of  them. 

Let  A  be  a  polyhedral  angle  con-  D 

tained  by  the  three  plane  angles 
BAC,  CAD,  DAB,  and  let  BAC  be 
the  greatest  of  these  angles. 

Then  CAD  +  DAB  >  BAC. 

For,  in  the  plane  BAC,  draw  the 
straight  line  AE,  making  the  angle 
BAE  =  angle  BAD. 


GEOMETRY. 


Make,  also,  AE  =  AD,  and  through  E  draw  any 
straight  line  BEC,  cutting  AB,  AC  in  the  points  B,  C ; 
join  D,  B  ;  D,  C  ; 

Then,  because  AD  =  AE,  and  AB  is  common  to 
the  two  triangles  DAB,  BAE,  and  the  angle  DAB  = 
angle  BAE,  by  construction, 

.-.  BD  =  BE. 
But,  in  the  triangle  BDC, 

BD  -f  DC>  BE  +  EC  (ax.  13,  cor.), 
.-.  DC  >  EC. 
Again,  v*  AD  =  AE,  and  AC  is  common  to  the 
two  triangles  DAC,  EAC,  but  the  base  DC  >  base 
EC, 

.\  angle  DAC  >  angle  EAC  (th.  32). 
But  angle  DAB  =  angle  BAE  ; 

.*.  angle  CAD  -f  angle  DAB  >  angle  BAE  -f  angle 
EAC  >  angle  BAC. 


PROP.    II. 

The  sum  of  the  plane  angles  which  form  a  polyhedral 
angle  is  always  less  than  four  right  angles. 

Let  P  be  a  polyhedral  angle  con- 
tained by  any  number  of  plane  angles 
APB,  BPC,  CPD,  DPE,  EPA. 

Let  the  polyhedral  angle  P  be  cut 
by  any  plane  ABCDE. 

Take  any  point  O  in  this  plane ; 
join  A,  O  ;  B,  O ;  C,  O  ;  D,  O ;  E,  O. 

Then,  since  the  sum  of  all  the  an- 
gles of  every  triangle  is  always  equal 
to  two  right  angles,  the  sum  of  all  the  angles  of  the 

triangles  APB,  BPC, about  the  point  P,  will  be 

equal  to  the  sum  of  all  the  angles  of  the  equal  number 
of  triangles  AOB,  BOC, about  the  point  O. 

Again,  by  the  last  Prop.,  angle  ABC  <  angle  ABP 
+  angle  CBP ;  in  like  manner,  angle  BCD  <  angle 
BCP  +  DCP,  and  so  for  all  the  angles  of  the  polygon 
ABCDE. 

signifies  "  because." 


P0LYIIEDKAL    ANGLES.  3 

Hence  the  sum  of  the  angles  at  the  bases  of  the 
triangles  whose  vertex  is  O,  is  less  than  the  sum  of 
the  angles  at  the  bases  of  the  triangles  whose  vertex 
is  P. 

.*.  The  sum  of  the  angles  about  the  point  O  must 
be  greater  than  the  sum  of  the  angles  about  the  point 
P. 

But  the  sum  of  the  angles  about  the  point  O  is  four 
right  angles. 

.\  The  sum  of  the  angles  about  the  point  P  is  less 
than  four  right  angles. 

PROP.    III. 

If  two  trihedral  angles  he  formed  by  three  plane  an- 
gles which  are  equal,  each  to  each,  the  planes  in  which 
these  angles  lie  will  be  equally  inclineorto  each  other. 

Let  P,  Q  be  two  trihedral 
angles  ; 

Let  angle  APC  ==  angle 
DQF,  angle  APB  *=  angle 
DQE,  and  angle  BPC  =  an- 
gle EQF. 

Then  the  inclination  of  the 
planes  APC,  APB  will  be  equal  to  the  inclination  of 
the  planes  DQF,  DQE. 

Take  any  point  B  in  the  intersection  of  the  planes 
APB,  CPB. 

From  B  draw  BY  perpendicular  to  the  plane  APC, 
meeting  the  plane  in  Y. 

From  Y  draw  YA,  YC,  perpendiculars  on  PA,  PC; 
join  A,  B  ;  B,  C. 

Again,  take  QE  =  PB  ;  from  E  draw  EZ  perpendic- 
ular to  the  plane  DQF,  meeting  the  plane  in  Z ;  from 
Z  draw  ZD,  ZF,  perpendiculars  on  QD,  QF ;  join 
D,  E  ;  E,  F. 

BA  is  perpendicular  to  PA  (Geom.  of  Planes,  Prop. 
8),  and  the  triangle  PAB  is  right-angled  at  A,  and  the 
triangle  QDE  is  right-angled  at  D. 

Also,  the  angle  APB  =  angle  DQE,  by  hyp. 


4  GEOMETRY. 

Moreover,  the  side  PB  =  side  QE  (by  construction); 
.*.  the  two  triangles  APB,  DQE  are  identical  (cor.  8, 
th.  15). 

.\  PA-QD,  and  AB  =  DE. 

In  like  manner,  we  can  prove  that 
PC  =  QF,  andBC=EF. 

Let  now  the  angle  APC  be  placed  upon  the  equal 
angle  DQF,  then  the  point  A  will  fall  upon  the  point 
D,  and  the  point  C  on  the  point  F,  because  PA  =  QD, 
andPC  =  QF. 

At  the  same  time,  AY,  which  is  perpendicular  to 
PA,  will  fall  upon  DZ,  which  is  perpendicular  to  QD  ; 
and,  in  like  manner,  CY  will  fall  upon  FZ. 

Hence  the  point  Y  will  fall  on  the  point  Z,  and  we 
shall  have 

AY  =  DZ,  andCY  =  FZ. 

But  the  triangles  AYB,  DZE  are  right-angled  in 
Y  and  Z,  the  hypothenuse  AB  =  hypothenuse  DE, 
and  the  side  AY  =  side  DZ  ;  hence  these  two  tri- 
angles are  equal  (th.  26,  cor.  2). 

.-.  angle  YAB=  angle  ZDE. 

The  angle  YAB  is  the  inclination  of  the  planes  APC, 
APB  (Geom.  of  Planes,  def.  6) ;  and 

The  angle  ZDE  is  the  inclination  of  the  planes 
DQF,  DQE. 

.*.  These  planes  are  equally  inclined  to  each  other. 

In  the  same  manner,  we  prove  the  angle  YCB  = 
angle  ZFE,  and,  consequently,  the  inclination  of  the 
planes  APC,  BPC  is  equal  to  the  inclination  of  the 
planes  DQF,  EQF. 

We  must,  however,  observe  that  the  angle  A,  of 
the  right-angled  triangle  YAB,  is  not,  properly  speak- 
ing, the  inclination  of  the  two  planes  APC,  APB,  ex- 
cept when  the  perpendicular  BY  falls  upon  the  same 
side  of  PA  as  PC  does  ;  if  it  fall  upon  the  other  side, 
then  the  angle  between  the  two  planes  will  be  obtuse, 
and,  added  to  the  angle  A  of  the  triangle  YAB,  will 
make  up  two  right  angles.  But,  in  this  case,  the  an- 
gle between  the  two  planes  DQF,  DQE  will  also  be 
obtuse,  and,  added  to  the  angle  D  of  the  triangle  ZDE, 
will  make  up  two  right  angles. 


POLYHEDRAL    ANGLES.  5 

Since,  then,  the  angle  A  will  always  be  equal  to  the 
angle  D,  we  infer  that  the  inclination  of  the  two 
planes  APC,  APB  will  always  be  equal  to  the  in- 
clination of  the  two  planes  DQF,  DQE.  In  the  first 
the  inclination  of  the  plane  is  the  angle  A  or  D  ; 
in  the  second  case,  it  is  the  supplement  of  those  angles. 

Scholium.  If  two  trihedral  angles  have  the  three 
plane  angles  of  the  one  equal  to  the  three  plane  an- 
gles of  the  other,  each  to  each,  and,  at  the  same  time, 
the  corresponding  angles  arranged  in  the  same  manner 
in  the  two  trihedral  angles,  then  these  two  trihedral 
angles  will  be  equal ;  and  if  placed  one  upon  the  other, 
they  will  coincide.  In  fact,  we  have  already  seen 
thai  the  quadri lateral  PAYC  will  coincide  with  the 
quadrilateral  QDZF.  Thus  the  point  Y  falls  upon 
the  point  Z,  and.  in  consequence  of  the  equality  of 
the  triangles  AYB,  DZE,  the  straight  line  YB,  per- 
pendicular to  the  plane  APC,  is  equal  to  the  straight 
line  ZE,  perpendicular  to  the  plane  DQE  ;  moreover, 
these  perpendiculars  lie  in  the  same  direction  ;  hence 
the  point  B  will  fall  upon  the  point  E,  the  straight 
line  PB  on  the  straight  line  QE  (their  extremities  al- 
ready coinciding),  and  the  .two  trihedral  angles  will 
entirely  coincide  with  each  other. 

This  coincidence,  however,  can  not  take  place  ex- 
cept we  suppose  the  equal  plane  angles  to  be  arrang- 
ed in  the  same  manner  in  the  two  trihedral  angles  ; 
for  if  the  equal  plane  angles  be  arranged  in  an  inverse 
order,  or,  which  comes  to  the  same  thing,  if  the  per- 
pendiculars YB,  ZE,  instead  of  being  situated  both  on 
the  same  side  of  the  planes  APC,  DQF,  were  situated 
on  opposite  sides  of  these  planes,  then  it  would  be 
impossible  to  make  the  two  trihedral  angles  coincide 
with  each  other.  It  would  not,  however,  be  less  true, 
according  to  the  above  theorem,  that  the  planes,  in 
which  the  equal  angles  lie,  would  be  equally  inclined 
to  each  other;  so  that  the  two  trihedral  angles  would 
be  equal  in  all  their  constituent  parts,  without  admit- 
ting of  superposition.  This  species  of  equality  is 
termed  symmetry. 


GEOMETRY. 


Thus  the  two  trihedral  angles  in  question,  which 
have  the  three  plane  angles  of  the  one  equal  to  the 
three  plane  angles  of  the  other,  each  to  each,  hut  ar- 
ranged in  an  inverse  order,  are  termed  angles  equal  by 
symmetry,  or,  simply,  symmetrical  angles. 

The  same  observation  applies  to  polyhedral  anles 
formed  by  more  than  three  plane  angles.  Thus,  a 
polyhedral  angle  formed  by  the  plane  angles  A,  B,  C, 
D,  E,  and  another  polyhedral  angle  formed  by  the 
same  angles  in  an  inverse  order,  A,  E,  D,  C,  B,  may 
be  such  that  the  planes  in  which  the  equal  angles  are 
situated  are  equally  inclined  to  each  other.  These 
two  polyhedral  angles,  which  would  in  this  case  be 
equal,  although  not  admitting  of  superposition,  would 
be  termed  polyhedral  angles  equal  by  symmetry,  or 
symmetrical  polyhedral  angles. 

In  plane  figures  there  is  no  species  of  equality  to 
which  this  designation  can  belong,  for  all  those  cases 
to  which  the  term  might  seem  to  apply  are  cases  of 
absolute  equality,  or  equality  of  coincidence.  The 
reason  of  this  is,  that  in  a  plane  figure  one  may  take 
the  upper  part  for  the  under,  and  vice  versa.  This, 
however,  does  not  hold  in  solids,  in  which  the  third 
dimension  may  be  taken  in  two  different  directions. 

This  term  symmetrical  is  of  very  extensive  appli- 
cation. Two  magnitudes  are  said  to  be  symmetrical 
with  respect  to  a  plane  when  the  corresponding  points 
are  on  opposite  sides  of  the  plane  in  the  same  per- 
pendicular to  it,  and  at  equal  distances  from  it. 

Thus  the  two  halves  of  the  human  body  are  sym- 
metrical with  respect  to  what  anatomists  call  the 
median  plane.     (See  Appendix  V.) 

A  plane  figure  may  be  said  to  be  symmetrical  with 
respect  to  a  median  line  when  points  on  one  side  of 
the  median  line  are  at  equal  perpendicular  distances 
from  it  with  opposite  points  on  the  other  side  (see 
Appendix  II.,  Del*.  2). 


EXERCISES.  7 

EXERCISES. 

1.  To  make  a  trihedral  angle  with  three  given  plane  angles. 

2.  Trove  that  iu  a  trihedral  angle  the  sum  of  the  diedral  angles  is 
greater  than  two  and  less  than  six  right  angles. 

3.  That  two  trihedral  angles  are  equal  when  they  have  two  plane 

and  the  included  diedral  angle  equal  [disposed  in  the  same 
order]. 

4.  Also,  when  they  have  one  plane  angle  and  two  adjacent  diedral 
angles. 

5.  Also,  when  they  have  three  diedral  angles  equal. 

6.  Show  that  if  from  a  point  within  a  trihedral  angle  perpendicu- 
lars be  drawn  to  each  of  the  planes  which  compose  it,  a  new  trihedral 
will  be  formed  whose  plane  angles  will  be  supplements  of  the  diedral 
angles  of  the  first,  and  vice  versa. 

7.  Prove  that  the  three  planes  bisecting  the  diedral  angles  of  a  tri- 


SOLID  GEOMETRY. 


DEFINITIONS. 

1.  A  Polyhedron  is  a  solid  bounded  by  planes. 
The  intersection  of  any  two  of  the  pjanes  is  called  a 
side  or  edge  of  the  polyhedron.  Each  bounding 
plane  will  be  a  polygon,  and  is  called  a  face  of  the 
polyhedron. 

2.  Similar  polyhedrons  are  such  as  have  all  their 
solid  angles  equal,  each  to  each,  and  are  contained  by 
the  same  number  of  similar  planes.* 

3.  A  Pyramid  is  a  solid  figure  contained 
by  triangular  planes  meeting  in  one  point, 
called  the  Vertex,  and  terminating  in  the 
sides  of  a  polygon,  called  the  Base  of  the 
pyramid. 

A  Regular  Pyramid  is  one  the  base  of  which  is  a 
regular  polygon,  and  the  vertex  in  a  perpendicular  to 
the  base  at  its  center. 

4.  A  Prism  is  a  solid  figure  contained  by 
plane  figures,  of  which  two  that  are  oppo- 
site are  equal,  similar,  and  parallel  to  each* 
other,  called  bases  ;  and  the  others  are  par- 
allelograms. The  latter  are  together  called 
the  Lateral  Surface  of  the  prism. 

A  Right  Prism  is  one  in  which  the  parallelograms 
are  perpendicular  to  the  bases. 

Pyramids  and  prisms  are  called  Triangular,  Quad- 
rangular, Pentagonal,  &c,  according  as  their  base  is 
a  triangle,  quadrilateral,  pentagon,  &c. 

5.  A  Sphere  is  a  solid  figure  described  by  the  rev- 
olution  of  a  semicircle  about  its  diameter,  which  re- 
mains unmoved.  The  moving  semicircle  is  called 
the  generatrix. 

6.  The  Axis  of  a  sphere  is  the  fixed  right  line  about 
which  the  semicircle  revolves. 

*  For  a  more  comprehensive  definition  of  similar  solids,  see  Ap- 
pendix V. 


GEOMETRY. 


7.  The  Center  of  a  sphere  is  the  same  with  that  of 
the  semicircle. 

8.  The  Diameter  of  a  sphere  is  any  right  line  which 
passes  through  the  center,  and  is  terminated  both 
ways  by  the  superficies  of  the  sphere.  The  axis  is  a 
diameter. 

9.  A  right  Cone  is  a  solid  figure  described  by  the 
revolution  of  a  right-angled  triangle  about  one  of  the 
sides  containing  the  right  angle,  which  side  remains 
fixed. 

Thus  the  side  AC,  revolving  round  A 

AB,  one  of  the  sides  which  contains  the 
right  angle  and  remains  fixed,  gener- 
ates a  cone. 

If  the  fixed  side  be  equal  to  the  other 
side  containing  the  right  angle,  the 
cone  is  called  a  right-angled  cone  ;  if 
it  be  less  than  the  other  side,  an  obtuse- 
angled  ;  and  if  greater,  an  acute-an- 
gled cone. 

10.  The  Axis  of  a  cone  is  the  fixed  right  line  about 
which  the  triangle  revolves. 

In  the  figure  above,  AB  is  the  axis. 

The  moving  side  of  the  triangle  is  called  the  gen- 
eratrix o*f  the  cone,  and  any  one  of  the  positions  of 
the  generatrix  is  called  an  element  of  the  cone.  The 
length  of  the  element  is  called  the  apophthegm  of  the 
cone. 

11.  The  Base  of  a  cone  is  the  circle  described  by 
that  side  containing  the  right  angle  which  revolves. 

12.  A  Cylinder  is  a  solid  figure  described  by  the 
revolution  of  a  right-angled  parallelogram  about  one 
of  its  sides  which  remains  fixed. 

Thus,  the  revolution  of  the  parallelo- 
gram AC  about  its  side  AB,  which  re- 
mains fixed,  generates  a  cylinder.* 

13.  The  axis  of  a  cylinder  is  the  fixed 
right  line  about  which  the  parallelogram 
revolves. 

*  The  solids  above  defined  are  properly  right  cones  and  cylinders 
These  may  also  be  oblique. 


SOLID    GEOMETRY. 


The  Altitude  of  a  pyramid  or  cone  is  the  perpen- 
dicular let  fall  from  the  vertex  to  the  plane  of  the 
base,  produced  if  necessary. 

The  altitude  of  a  prism  or  cylinder  is  the  perpen- 
dicular distance  between  the  parallel  bases. 

The  altitude  of  a  cone  or  cylinder  is  identical  with 
the  axis. 

14.  The  bases  of  a  cylinder  are  the  circles  de- 
scribed by  the  two  revolving  opposite  sides  of  the 
parallelogram.* 

15.  Similar  cones  and  cylinders  are  those  which 
have  their  axes  and  the  diameters  of  their  bases  pro- 
portionals. 

16.  A  regular  polyhedron  is  one,  all  whose  solid 
angles  are  equal,  and  whose  faces  are  equal  polygons. 

17.  A  Cube  is  a  regular  solid  figure  con- 
tained by  six  equal  squares. 


18.  A  regular  Tetrahedron  is  a  solid 
figure  contained  by  four  equal  and  equi- 
lateral triangles. 


10.  A  regular  Octahedron  is  a  solid 
figure  contained  by  eight  equal  and  equi- 
lateral triangles. 


20.  A  regular  Dodecahedron  is  a  solid 
figure  contained  by  twelve  equal  penta- 
gons which  are  equilateral  and  equian- 
gular. 


*  Every  section  of  a  cylinder  made  by  a  plane  perpendicular  to  the 
axis  is  a  circle,  and  every  section  through  the  axis  is  a  rectangle 
double  the  generating  rectangle. 

The  moving  side  of  the  parallelogram  is  called  the  generatrix  of  the 
cylinder,  and  any  one  of  its  positions  is  called  an  element. 


GEOMETRY. 


21.  An  Icosahedron  is  a  solid  figure 
contained  by  twenty  equal  and  equilateral 
triangles. 


These  five,  it  will  be  shown  hereafter,  are  the  only 
regular  polyhedrons  which  can  be  formed. 

22.  A  Parallelopipedon  is  a  solid  figure  con- 
tained by  six  parallelograms,  the  planes  of  ev- 
ery opposite  two  whereof  are  parallel.  The 
parallelopipedon  is  a  prism  with  parallelo- 
grams for  bases. 


PROPOSITIONS. 


PROP. 


If  a  p?'is?n  be  cut  by  a  plane  parallel  to  its  base,  the 
section  will  be  equal  and  like  to  the  base. 

Let  AG  be  any  prism,  and  IL  a  plane        H    G 
parallel  to  the  base  AC  ;  then  will  the 
plane  IL  be  equal,  and  like  to  the  base 
AC,  or  the  two  planes  will  have  all  their 
sides  and  all  their  angles  equal. 

For  the  two  planes  AC,  IL,  being  par- 
allel, by  hypothesis,  and  two  parallel 
planes,  cut  by  a  third  plane,  having  par- 
allel sections  ;  therefore,  IK  is  parallel  to  AB,  KL  to 
BC,  LM  to  CD,  and  IM  to  AD.  But  AI  and  BK  are 
parallels,  by  def.  4,  last  page  but  two ;  consequently, 
AK  is  a  parallelogram  ;  and  the  opposite  sides,  AB, 
IK,  are  equal.  In  like  manner,  it  is  shown  that  KL 
is  =BC  and  LM  =  CD,  and  IM  =  AD,  or  the  two 
planes  AC,  IL  are  mutually  equilateral.  But  these 
two  planes,  having  their  corresponding  sides  parallel, 
have  the  angles  contained  by  them  also  equal  (Geom. 
of  Planes,  Prop.  16)  :  namely,  the  angle  A  =  the  angle 
I,  the  angle  B  =  the  angle  K,  the  angle  C  =  the  angle 


SOLID    GEOMETRY. 


L,  and  the  angle  D  —  the  angle  M.  So  that  the  two 
planes  AC, IL  have  all  their  corresponding  sides  and 
angles  equal,  or  are  equal  and  like.     Q.  E.  D. 


PROP.   II. 


If  a  cylinder  be  cut  by  a  plane  parallel  to  its  base, 
the  section  ivill  be  a  circle  equal  to  the  base. 

Let  AF  be  a  cylinder,  and  GHI  any 
section  parallel  to  the  base  ABC  ;  then  D 
will  GHI  be  a  circle  equal  to  ABC. 

For,  let  the  plane  KE  pass  through  the 
axis  of  the  cylinder  MK,  and  meet  the 
section  GHI  in  the  line  LH. 

Then,  since  KL,  BH  are  parallel  (def. 
12,  Sol.Geom.) ;  and  the  plane  KH  meet- 
ing the  two  parallel  planes  ABC,  GHI,  makes  the  two 
sections  KB,  LH  parallel  (Prop.  12,  Geom.  of  Planes) ; 
the  figure  KLHB  is,  therefore,  a  parallelogram,  and, 
consequently,  has  the  opposite  sides  LH,  KB  equal, 
where  KB  is  a  radius  of  the  circular  base. 

In  like  manner,  it  may  be  shown  that  any  other  line 
drawn  from  the  point  L  to  the  circumference  of  the 
section  GHI,  is  equal  to  the  radius  of  the  base  ;  con- 
sequently, GHI  is  a  circle,  and  equal  to  ABC.  Q. 
E.  D. 


/ 
/ 
/ 

— 

/ 

/ 

/" 

PROP.   III. 

All  prisms,  and  a  cylinder,  of  equal  bases  and  alti- 
tudes, are  equal  to  each  other. 

Let  AC,  DF  be  two 

prisms  and  a  cylinder, 
upon  equal  bases  AB, 
DE,  and  having  equal 
altitudes;  then  will  the 
solids  AC,  DF  be  equal. 
For,  let  PQ,  RS  be 
any  two  sections  parallel  to  the  bases,  and  equidistant 
from  them.     Then,  by  the  last  two  propositions,  the 

G 


v 


GEOMETRY 


section  PQ,  is  equal  to  the  base  AB,  and  the  section 
RS  equal  to  the  base  DE.  But  the  bases  AB,  DE  are 
equal  by  the  hypothesis  ;  therefore  the  sections  PQ, 
RS  are  also  equal.  And  in  like  manner  it  may  be 
shown  that  any  other  corresponding  sections  are 
equal  to  one  another. 

Since,  then,  every  section  in  the  prism  AC  is  equal 
to  its  corresponding  section  in  the  prism,  or  cylinder 
RS,  the  prisms  and  cylinder  themselves,  which  are 
composed  of  those  sections  (which  will  be  the  same 
in  number,*  the  altitudes  being  equal),  must  also  be 
equal.     Q.  E.  D. 

Corol.  Every  prism,  or  cylinder,  is  equal  to  a  rect- 
angular parallelopipedon,  of  an  equal  base  and  alti- 
tude. 


S    G 


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A     L     M    N 


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H     i 

OF 

W 

PROP.  IV. 

Rectangular  parallelopipedons,  of  equal  altitudes, 
have  to  each  other  the  same  proportion  as  their  bases. 

Let  AC,  EG  be  two  rectan-  q  r 
gular  parallelopipedons,  hav- 
ing the  equal  altitudes  AD, 
EH ;  then  will  AC  be  to  EG 
as  the  base  AB  is  to  the  base 
EF. 

For,  let  the  proportion  of 
the  base  AB  to  the  base  EF  be  that  of  any  one  num- 
ber m  (3)  to  any  other  number  n  (2).  And  conceive 
AB  to  be  divided  into  m  equal  parts,  or  rectangles, 
AI,  LK,  MB  (by  dividing  AN  into  that  number  of 
equal  parts,  and  drawing  IL,  KM  parallel  to  BN). 
And  let  EF  be  divided,  in  like  manner,  into  n  equal 
parts,  or  rectangles  EO,  PF :  all  of  these  parts  of 
both  bases  being  mutually  equal  among  themselves. 
And  through  the  lines  of  division  let  the  plane  sec- 
tions LR,  MS,  PV  pass  parallel  to  AQ,  ET. 


e    P 


*  The  number  in  each  will  be  infinite,  but  these  infinities  will  evi- 
dently be  equal. 


SOLID    GEOMETRY.  7 

Then  the  parallelopipedons  AR,  LS,  MC,  EV,  PG 
are  all  equal,  having  equal  bases  and  heights.  There- 
fore, the  solid  AC  is  to  the  solid  EG  as  the  number  of 
parts  in  AC  to  the  number  of  equal  parts  in  EG,  or 
as  the  number  of  parts  in  AB  to  the  number  of  equal 
parts  in  EF ;  that  is,  as  the  base  AB  to  the  base  EF. 
Q.  E.  D. 

Note.  If  the  bases  be  incommensurable,  the  divisions 
must  be  infinitely  small. 

Corol.  From  this  proposition,  and  the  corollary  to 
the  last,  it  appears  that  all  prisms  and  cylinders  of 
equal  altitudes  are  to  each  other  as  their  bases  ;  every 
prism  and  cylinder  being  equal  to  a  rectangular  par- 
allelopipedon  of  an  equal  base  and  height. 

prop.  v. 

Rectangular  parallelopipedons  of  equal  bases  are  in 
proportion  to  each  other  as  their  altitudes. 

Let  AB,  CD  be  two  rect- 
angular parallelopipedons 
standing  on  the  equal  bases 
AE,  CF ;  then  will  AB  be  to 
CD  as  the  altitude  EB  is  to 
the  altitude  DF. 

For,  let  AG  be  a  rectangu- 
lar parallopipedon  on  the  base  AE,  and  its  altitude 
EG  equal  to  the  altitude  FD  of  the  solid  CD. 

Then  AG  and  CD  are  equal,  being  prisms  of  equal 
bases  and  altitudes.  But  if  HB,  HG  be  considered 
as  bases,  the  solids  AB,  AG- of  equal  altitude  AH, 
will  be  to  each  other  as  those  bases  HB,  HG.  But 
these  bases  HB,  HG  being  parallelograms  of  equal 
altitude  HE,  are  to  each  other  as  their  bases  EB,  EG; 
and,  therefore,  the  two  prisms  AB,  AG  are  to  each 
other  as  the  lines  EB,  EG.  But  AG  is  equal  CD,  and 
EG  equal  FD  ;  consequently,  the  prisms  AB,  CD  are 
to  each  other  as  their  altitudes  EB,  FD  ;  that  is,  AB  : 
CD::EB:FD.     Q.  E.  D. 

Corol.  From  this  proposition  and  the  corollary  to 


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H 

JE 

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.  1 

8 


GEOMETRY. 


Prop.  3,  it  appears  that  all  prisms  and  cylinders  of 
equal  bases  are  to  one  another,  as  their  altitudes. 


PROP.  VI. 

Rectangular  parallelopipedons  are  to  each  other  as 
the  products  of  their  bases  by  their  altitudes. 

.  TheparallelopipedonAF  B 

is  to  the  parallelopipedon 
CE  as  the  base  AG  x  the 
altitude  GF,  is  to  the  base 
CD  X  altitude  DE. 

For  the  parallelopipedons 
AB  and  CE,  having  the 
same  altitude,  are  to  each 
other  as  their  bases  AG  and 
CD;  and  the  parallelopi- 
pedons AF  and  AB,  having 

the  same  base,  are  to  each  A  D 

other  as  their  altitudes  GF,  GB ;  or, 
AB:CE::AG:CD, 
AF:AB::GF:GB. 
Multiplying  these  two  proportions  together  and 
striking  out  AB  from  the  two  terms  of  the  first  result- 
ing ratio,  we  have 

AF :  CE  : :  AG  X  GF :  CD  X  GB.* 


/ 

/ 

f~ 

/ 

/ 

0 

/ 
/ 

E 

* 

G- 

/ 

/ 

*  As  rectangular  parallelopipeds  are  always  to  each  other  as  the 
products  of  their  bases  by  their  altitudes,  this  may  be  taken  as  the 
measure  of  parallelopipeds;  and  as  every  parallelopiped  is  equal  to  a 
prism  or  cylinder  of  the  same  base  and  altitude  (Prop.  3),  it  follows 
that  the  measure  of  any  prism  or  cylinder  is  the  product  of  its  base 
by  its  altitude. 

If  the  convex  surface  of  a  cylinder  be  developed,  it  opens  out  into 
a  parallelogram,  of  which  the  circumference  of  the  cylinder's  base  is 
the  base,  and  the  altitude  of  which  is  that  of  the  cylinder;  and  as  this 
parallelogram  is  measured  by  the  product  of  its  base  by  its  altitude, 
we  have  for  the  measure  of  the  convex  surface  of  a  cylinder  the  prod- 
uct of  the  circumference  of  its  base  by  its  altitude. 

A  plane  determined  by  an  element  of  a  cylinder,  and  the  tangent 
line  to  the  base  at  the  point  where  the  element  meets  it,  is  a  tangent 
plane  to  the  cylinder. 

The  contact  is  along  the  whole  length  of  the  element,  which  i* 
called  the  element  of  contact. 


SOLID    GEOMETRY. 


PROP.    VII. 


Similar'  prisms  and  cylinders  are  to  each  other  as 
the  cubes  of  their  altitudes,  or  of  any  other  like  linear 
dimensions. 

Let  ABCD,  EFGH  be  two         /x 

similar  prisms  ;  then  will  the  D^        / 


\ 


"9 


<& 


prism  CD  be  to  the  prism  GH 
as  AB3  to  EF3,  or  as  AD3  to 
EH'. 

For  the  solids  are  to  each 
other  as  the  products  of  their 
bases  and  altitudes  (by  the  note        a  r 

to  the  last  Prop.),  that  is,  as  AC .  AD  to  EG .  EH. 
But  the  bases  being  similar  planes,  are  to  each  other 
as  the  squares  of  their  like  sides,  that  is,  AC  to  EG 
as  ABa  to  EF3 ;  therefore,  substituting  the  ratio  or 
fraction  AB2 :  EF3  for  AC  :  EG,  we  have  the  solid  CD 
to  the  solid  GH  as  AB2 .  AD  to  EF3 .  EH.  But  BD 
and  FH  being  similar  planes,  have  their  like  sides 
proportional,  that  is,  AB:  EF  : :  AD  :  EH,  or  AB9 : 
EF3 : :  AD3 :  EH3 ;  multiply  this  by  the  identical  pro- 
portion AD  :  EH  : :  AD  :  EH,  and  we  have  AB3 .  AD 
:  EF3 .  EH  : :  AD3 :  EHJ ;  and,  consequently,  the  solid 
CD  :  solid  GH : :  AD3 :  EH3,  or  AB3 :  EFa.     Q.  E.  D. 


PROP.   VIII. 

In  a  pyramid,  a  section  parallel  to  the  base  is  similar 
to  the  base,  and  these  two  planes  will  be  to  each  other  as 
the  squares  of  their  distances  from  the  vertex. 

Let  ABCD  be  a  pyramid,  and  EFG  a  A 
section  parallel  to  the  base  BCD,  also  AIH 
a  line  perpendicular  to  the  two  planes  at 
H  and  I ;  then  will  BD,  EG  be  two  simi- 
lar planes,  and  the  plane  BD  will  be  to  the 
plane  EG  as  AH3  to  AP. 

For,  join   CH,  FI.     Then,  because   a 
plane  cutting  two  parallel  planes  makes  B 


10  GEOMETRY. 

parallel  sections,  therefore  the  plane  ABC,  meeting  the 
two  parallel  planes  BD,  EG,  makes  the  sections  BC, 
EF  parallel ;  in  like  manner,  the  plane  ACD  makes  the 
sections  CD,  FG  parallel.  Again,  because  two  pair  of 
parallel  lines  make  equal  angles,  the  two  EF,  FG, 
which  are  parallel  to  BC,  CD,  make  the  angle  EFG 
equal  the  angle  BCD.  And,  in  like  manner,  it  is 
shown  that  each  angle  in  the  plane  EG  is  equal  to 
each  angle  in  the  plane  BD,  and,  consequently,  those 
two  planes  are  equiangular. 

Again,  the  three  lines  AB,  AC,  AD,  making  with 
the  parallels  BC,  EF,  and  CD,  FG,  equal  angles;  and 
the  angles  at  A  being  common,  the  two  triangles 
ABC,  AEF  are  equiangular,  as  also  the  two  triangles 
ACD,  AFG,  and  have,  therefore,  their  like  sides 
proportional,  namely,  AC  :  AF : :  BC  :  EF : :  CD :  FG. 
And,  in  like  manner,  it  may  be  shown  that  all  the  lines 
in  the  plane  EG  are  proportional  to  all  the  correspond- 
ing ones  in  the  base  BD.  Hence  these  two  planes, 
having  their  angles  equal  and  their  sides  proportional, 
are  similar. 

But  similar  planes  being  to  each  other  as  the  squares 
of  their  like  sides,  the  plane  BD  :  EG  : :  BC2 :  EF2 :  or 
: :  AC2 :  AF2,  by  what  is  shown  above.  But  the  two 
triangles  AHC,  AIF,  having  the  angles  H  and  I  right 
ones,  and  the  angle  A  common,  are  equiangular,  and 
have,  therefore,  their  like  sides  proportional,  namely, 
AC:AF::AH:AI,  or  AC2 :  AF2 : :  AH2 :  AI2.  Con- 
sequently, the  two  planes  BD,  EG,  which  are  as  the 
former  squares  AC2,  AF2,  will  be  also  as  the  latter 
squares  AH2,  AF,  that  is,  BD  :  EG  : :  AH2 :  AI2. 


PROP.  IX. 

In  a  right  cone  a  section  parallel  to  the  base  is  a  cir- 
cle,  and  this  section  is  to  the  base  as  the  squares  of 
their  distances  from  the  vertex. 

Let  ABCD  be  a  right  cone,  and  GHI  a  section 
parallel  to  the  base  BCD ;  then  will  GHI  be  a  circle, 


SOLID   GEOMETRY 


11 


and  BCD,  GHI  will  be  to  each  other 
as  the  squares  of  their  distances  from 
the  vertex. 

For,  let  the  planes  ACE,  ADE  pass 
through  the  axis  of  the  cone  AKE,  meet- 
ing t  he  section  in  the  three  points  H,  I,  K. 

Then,  since  the  section  GHI  is  paral- 
lel to  the  base  BCD,  and  the  planes  CK, 
DK  meet  them,  HK  is  parallel  to  CE, 
and  IK  to  DE.  And  from  similar  trian- 
gles, shown  to  be  such  (as  in  the  last  Prop.),  KH : 
EC  : :  AK  :  AE  : :  KI :  ED.  But  EC  is  equal  to  ED, 
being  radii  of  the  same  circle ;  therefore,  KI  is  also 
equal  to  KH.  And  the  same  may  be  shown  of  any 
other  lines  drawn  from  the  point  K  to  the  circumfe- 
rence of  the  section  GHI,  which  is,  therefore,  a  circle. 

Again,  since  AK  :  AE  : :  KI :  ED,  hence  AK2 : 
AE3 : :  KP  :  ED3 ;  but  KP  :  ED3 : :  circle  GHI :  circle 
BCD  (th.  72) ;  therefore,  AK2 :  AE3 : :  circle  GHI :  cir- 
cle BCD.     Q.  E.  D. 

PROP.  X. 

All  pyramids  and  right  cones  of  equal  bases  and  al- 
titudes are  equal  to  one  another. 

Let  ABC,  DEF 

be  any  pyramids 
and  a  cone,  of  equal 
bases  BC,  EF,  and 
equal  altitudes  AG, 
DH  ;  then  will  the 
pyramids  and  cone 
ABC  and  DEF  be 
equal. 

For,  parallel  to  the  bases,  and  at  equal  distances, 
AN,  DO,  from  the  vertices,  suppose  the  planes  IK, 
LM  to  be  drawn. 

Then,  by  Prop.  8  and  9, 

DO3 :  DH3 : :  LM :  EF, 
and  AN3:  AG3::  IK   :  BC. 

But,  since  AN2,  AG2  are  equal  to  DO2,  DH3 ;  there- 


12  GEOMETRV. 

fore  IK :  BC  : :  LM  :  EF.  But  BC  is  equal  to  EF,  by 
hypothesis  ;  therefore  IK  is  also  equal  to  LM. 

In  the  same  manner,  it  is  shown  that  any  other  sec- 
tions, at  equal  distance  from  the  vertex,  are  equal  to 
each  other. 

Since,  then,  every  section  in  the  cone  is  equal  to 
the  corresponding  section  in  the  pyramids,  and  the 
heights  are  equal,  the  solids  ABC,  DEF,  composed  of 
those  sections,  must  be  equal  also.     Q,.  E.  D. 

PROP.  XI. 

Every  triangular  prism  may  be  divided  into  three 
equal  triangular  pyramids  of  the  same  base  and  alti- 
tude with  the  prism. 

Let  ABCDEF  be  a  prism. 

In  the  planes  of  the  three  sides  of  the 
prism,  draw  the  diagonals  BF,  BD,  CD. 
Then  the  two  planes  BDF,  BCD  divide 
the  whole  prism  into  the  three  pyramids 
BDEF,  DABC,  DBCF  ;  which  are  proved 
to  be  all  equal  to  one  another  as  follows  : 

Since  the  opposite  ends  of  the  prism 
are  equal  to  each  other,  the  pyramid  whose  base  is 
ABC  and  vertex  D,  is  equal  to  the  pyramid  whose 
base  is  DEF  and  vertex  B  (Prop.  10),  being  pyramids 
of  equal  base  and  altitude. 

But  the  latter  pyramid,  whose  base  is  DEF  and 
vertex  B,  may  be  considered  as  having  BEF  for  its 
base  and  D  for  its  vertex,  and  this  is  equal  to  the  third 
pyramid,  whose  base  is  BCF  and  vertex  D,  being 
pyramids  of  the  same  altitude  (since  they  have  the 
same  vertex  and  their  bases  are  in  the  same  plane) 
and  equal  bases  BEF,  BCF  (th.  19). 

Consequently,  all  the  three  pyramids  which  com- 
pose the  prism  are  equal  to  each  other,  and  each  pyr- 
amid is  the  third  part  of  the  prism,  or  the  prism  is 
triple  of  the  pyramid.     Q.  E.  D. 

Corol.  1.  Any  triangular  pyramid  is  the  third  part 
of  a  triangular  prism  of  the  same  base  and  altitude 
(this  follows  from  the  last  Prop.,  10). 


SOLID   <;i:hmi;trY.  13 

Coral.  2.  Every  pyramid,  whatever  its  figure  may 
be,  is  the  third  part  of  a  prism  of  the  same  base  and 
altitude.  This  follows  from  Props.  3  and  10,  or  may 
be  proved  by  dividing  the  given  prism  into  triangular 
prisms,  and  the  given  pyramid  into  triangular  pyra- 
mids, ail  having  a  common  altitude. 

Corol.  3.  Any  right  cone  is  the  third  part  of  a  cyl- 
inder, or  of  a  prism,  of  equal  base  and  altitude  ;  since 
it  has  been  proved  that  a  cylinder  is  equal  to  a  prism, 
and  a  cone  equal  to  a  pyramid,  of  equal  base  and  al- 
titude. 

Corol.  4.  The  measure  of  a  pyramid  or  cone  will 
be  the  product  of  its  base  by  the  third  of  its  altitude. 
(See  note  to  Prop.  G.) 

Scholium.  Whatever  has  been  demonstrated  of  the 
proportionality  of  prisms  or  cylinders  holds  equally 
true  of  pyramids  or  cones,  the  former  being  always 
triple  the  latter  when  they  have  the  same  base  and 
altitude ;  viz.,  that  similar  pyramids  or  cones  are  as 
the  cubes  of  their  like  linear  sides,  or  diameters,  or 
altitudes,  &c. 

The  tangent  plane  to  a  cone  is  analogous  to  that 
of  a  cylinder.  The  contact  is  a  right-lined  element. 
Every  tangent  plane  to  a  cone  passes  through  the 
vertex. 

The  surface  of  a  cone  developes  into  the  sector  of 
a  circle,  and  is  measured  by  the  circumference  of  the 
base  multiplied  by  half  the  apophthegm. 

A  pyramid  is  inscribed  in  a  cone  when  the  base  of 
the  pyramid  is  inscribed  in  that  of  the  cone,  and  they 
have  the  same  vertex. 

A  cone  is  a  pyramid  of  an  infinite  number  of  trian- 
gular faces. 


EXERCISES. 

1.  Prove  that  the  four  diagonals  of  a  parallelopipedon  meet  in  the 
same  point. 

2.  That  the  square  of  each  diagonal  of  a  rectangular  parallelopiped- 
on is  equal  to  the  sum  of  the  squares  of  its  three  edges. 


14  GEOMETRY. 

3.  Construct  a  parallelopipedon  upon  three  lines  perpendicular  to 
each  other  as  edges. 

4.  Prove  that  the  two  lines  joining  the  points  of  the  opposite  faces 
of  a  parallelopipedon,  in  which  the  diagonals  of  those  faces  intersect, 
bisect  each  other  at  the  point  where  the  diagonals  of  the  solid  meet. 

5.  Prove  that  two  polyhedrons  which  have  the  same  vertices  are 
identical. 

C.  That  two  prisms  are  equal  when  they  have  three  faces  forming 
a  polyhedral  angle  of  the  one  equal  to  the  same  in  the  other,  and  ar- 
ranged in  the  same  order. 

7.  That  two  right  prisms  are  equal  when  they  have  equal  bases 
and  altitudes. 

8.  That  two  tetrahedrons  are  equal,  1°.  When  they  have  a  diedral 
angle  equal,  comprehended  between  equal  faces,  arranged  in  the 
same  manner  in  both;  2°.  When  they  have  one  trihedral  angle, 
comprehended  by  three  equal  faces  in  each,  and  arranged  in  the  same 
order.  3°.  Corol.  When  they  have  their  edges  all  equal  and  arrang- 
ed in  the  same  order.  4°.  When  they  have  two  faces  and  two  adjacent 
diedrals  equal. 

9.  That  two  pyramids  are  equal  when  they  have  their  bases  and 
two  other  faces  forming  a  trihedral  angle,  with  the  base  equal  in  each. 

10.  Polyhedrons  may  be  divided  into  tetrahedrons  by  planes  pass- 
ing diagonally  through  the  edges. 

11.  Polyhedrons  are  equal  when  composed  of  the  same  number  of 
equal  tetrahedrons. 

12.  Prove  that  polyhedrons  are  equal  when  their  faces  and  diedral 
angles  are  equal,  and  disposed  in  the  same  order. 

13.  Prove  that  similar  polyhedrons  are  composed  of  the  same  num- 
ber of  similar  tetrahedrons. 

14.  That  polyhedrons  are  similar  when  their  faces  are  all  equal, 
each  to  each,  and  equally  inclined. 

15.  That  polyhedrons  are  similar  when  thsy  have  a  face  in  each 
similar,  and  their  homologous  vertices  out  of  this  face  are  determined 
by  tetrahedrons  having  a  triangular  face  in  the  homologous  face.* 

16.  That  two  pyi-amids  are  similar  when  they  have  their  edges 
parallel. 

17.  That  two  regular  polyhedrons  of  the  same  kind  are  similar. 

18.  That  a  plane  passed  through  two  edges  of  a  parallelopiped, 
diagonally  opposed  to  each  other,  divides  the  parallelopiped  into  two 
symmetrical  triangular  prisms  equal  in  volume. 

*  This  is  the  definition  of  similar  polyhedrons  given  by  Legendre. 


EXERCISES.  15 

19.  That  two  prisms  of  the  same  base  are  proportional  to  their 
altitudes. 

20.  That  two  regular  pyramids  are  equal  when  their  base  and  an 
edge  of  the  one  are  equal  to  the  same  in  the  other. 

21.  That  similar  pyramids  are  to  each  other  as  the  cubes  of  their 
homologous  sides. 

20.  That  similar  polyhedrons  are  as  the  cubes  of  their  homologous 
sides. 

23.  That  the  surfaces  of  similar  polyhedrons  are  as  the  squares  of 
their  homologous  sides. 

21.  Cut  a  pyramid  by  a  plane,  pai-allel  to  the  base,  in  such  a  way 
that  the  section  shall  be  to  the  base  in  the  ratio  of  two  given  lines. 

25.  Also,  so  that  the  convex  surface  of  the  superior  portion  shall  be 
one  third  that  of  the  whole  pyramid. 

26.  Show  how  to  construct  a  pyramid  when  the  base  and  two  ad- 
jacent triangular  faces  are  given. 

27.  A  prism  with  the  same  data. 

28.  A  parallelopipedon  with  given  base,  and  edge  meeting  it. 

29.  With  given  edges  to  construct  the  faces  of  a  tetrahedron. 

30.  With  three  edges  forming  a  trihedral  angle,  and  their  angles, 
to  construct  the  faces  of  a  triangular  prism. 

31.  The  same  for  a  pentagonal  prism,  three  faces,  forming  a  tri- 
hedral angle,  being  given. 

32.  Show  how  to  construct  a  cylinder  similar  to  a  given  cylinder, 
and  whose  base  shall  be  to  that  of  the  given  in  the  ratio  of  two  given 
lines. 

33.  Show  in  what  the  frustum  of  a  cone  develops,  and  what  is  the 
measure  of  its  surface. 

31.  Prove  that  every  plane  parallel  to  the  axes  of  a  cylinder  cuts  its 
surface  in  two  lines  parallel  to  the  axis. 

:r>.  That  if  a  circle  and  a  line  tangent  to  it  revolve  about  a  com- 
mon axis  passing  through  the  center  of  the  circle,  the  curve  of  contact 
of  the  cone  generated  by  the  line,  and  the  sphere  generated  by  the 
circle,  will  be  a  circle  whose  plane  is  perpendicular  to  the  axis. 

36.  That  similar  cones  and  cylinders  are  proportional,  their  surfaces 
to  the  squares,  and  their  volumes  to  the  cubes  of  their  altitudes,  ele- 
ments, diameters  of  bases,  or  any  homologous  lines. 


16 


GEOMETRY. 


prop.  xn. 


The  frustum  of  a  pyramid  is  composed  of  three  pyr- 
amids having  for  a  common  altitude  the  altitude  of  the 
frustum,  and  for  bases  the  upper  and  lower  base  of  the 
frustum  and  a  mean  proportional  between  them. 

Let   a    plane    be    passed  c 

through  the  three  points  a, 
C,  b;  it  cuts  off  a  triangular 
pyramid  having  acb  for  a 
base,  and  C  in  the  plane  of 
the  lower  base  of  the  frus- 
tum for  a  vertex.  This  is 
evidently  the  first  pyramid 
of  the  enunciation.  Again : 
suppose  a  plane  be  passed 
through  the  three  points  A, 
C,  /; ;  it  cuts  off  a  pyramid 
having  ABC  for  base  and  b 
for  vertex,  the  second  pyra- 
mid of  the  enunciation.  There  remains  the  pyramid 
CAba,  which  is  equal  to  the  pyramid  CAD6  (M3  be- 
ing drawn  parallel  to  « A),  having  the  same  vertex  C 
and  an  equal  base,  since  the  diagonal  Kb  bisects  the 
parallelogram  ADba.  This  last  pyramid  being  con- 
sidered as  having  its  vertex  at  b  and  its  base  ADC, 
has  the  altitude  of  the  frustum,  and  it  remains  to  show 
that  the  triangle  ADC,  which  is  its  base,  is  a  mean 
proportional  between  the  triangles  ABC  and  abc.  For 
this  purpose,  let  us  observe  that 

A  ABC  :  A  ADC  :  AB :  AD (1) 

because  they  have  a  common  vertex,  and,  therefore, 
are  to  each  other  as  their  bases,  which  are  in  the  same 
straight  line.     Also,  that 

A  ADC  :  A  abc :  AC  :  ac (2) 

because  the  angle  A=  the  angle  a  (corol.  1,  Prop. 

16,  Geom.  of  Planes),  and  (th.  60,  cor.  3)  A  ADC : 

abc::  AD  X  AC  :ab  X  ac,  and  AD  =  «&.     Also,  that 

AB:ab::Adac (3) 


SOLID    GEOMETRY.  17 

since  the  triangles  ABC,  abc  are  similar  (Prop.  8, 
ante). 

►Substituting  now  the  first  ratio  of  (3)  for  its  equiv- 
alent, the  second  ratio  of  (2),  and  then  the  first  ratio 
of  (2)  for  its  equivalent  (since  AD  =ab),  the  second 
ratio  of  (I),  we  have 

A  ABC  :  a  ADC  : :  A  ADC  :  A  abc.     Q.  E.  D. 

CoroL  The  same  proposition  is  true  of  the  frustum 
of  any  pyramid  or  of  a  cone  (Prop.  9  and  10)  which 
is  equivalent  to  three  cones  having  the  upper  base, 
the  lower  base  and  a  mean  proportional  between  the 
two  for  bases,  and  for  a  common  altitude  the  altitude 
pf  the  frustum.  In  symbols  r  and  rf,  being  the  radii 
of  the  bases,  and  h  the  altitude  of  the  Irustum,  its 
volume  would  be  expressed  by  (th.  73,  schol.) 
n(?'u  +  r'a  +  rr')/i. 

TROP.    XIII. 

If  a  sphere  be  cut  by  a  plane,  the  section  will  be  a 
circle. 

Because  the  radii  of  the  sphere  are  all  equal,  each 
of  them  being  equal  to  the  radius  of  the  describing 
semicircle,  it  is  evident  that  if  the  section  pass  through 
the  center  of  the  sphere,  then  the  distance  from  the 
center  to  every  point  in  the  periphery  of  that  section 
will  be  equal  to  the  radius  of  the  sphere,  and  the  sec- 
tion will,  therefore,  be  a  circle  of  the  same  radius  as 
the  sphere.  But  if  the  plane  do  not  pass  through  the 
center,  draw  a  perpendicular  to  it  from  the  center, 
and  draw  any  number  of  radii  of  the  sphere  to  the 
intersection  of  its  surface  with  the  plane  ;  then  these 
radii  are  evidently  the  hypothenuses  of  a  correspond- 
ing number  of  right-angled  triangles,  which  have  the 
perpendicular  from  the  center  on  the  plane  of  the 
section,  as  a  common  side  ;  consequently,  their  other 
sides  are  all  equal,  and,  therefore,  the  section  of  the 
sphere  by  the  plane  is  a  circle,  whose  center  is  the 
point  in  which  the  perpendicular  cuts  the  plane. 

Scholium,  All  the  sections  through  the  center  are 


£*v 


II 


18  GEOMETRY. 

equal  to  one  another,  and  are  greater  than  any  other 
section  which  does  not  pass  through  the  center.  Sec- 
tions through  the  center  are  called  great  circles,  and 
the  other  sections  small  or  less  circles. 

PROP.  XIV. 

Every  sphere  is  two  thirds  of  its  circumscribing  cyl- 
inder. 

Let  ABCD  be  a  section  of  the  cyl-     A         F        n 
inder,  and  EFGH  a  section  of  the 
sphere  through  the  center  I,  and  join      /\ 
AI,  BI.     Let  FIH  be  parallel  to  AD  uf 
or  BC,  and  EIG  and  KL  parallel  to  E 
AB  or  DC,  the  base  of  the  cylindric 
section ;  the  latter  line  KL  meeting 
BI  in  M,  and  the  circular  section  of 
the  sphere  in  N. 

Then,  if  the  whole  plane  HFBC  be  conceived  to 
revolve  about  the  line  HF  as  an  axis,  the  square  FG 
will  describe  a  cylinder  AG,  and  the  quadrant  IFG 
will  describe  a  hemisphere  EFG,  and  the  triangle  IFB 
will  describe  a  cone  IAB.  Also,  in  the  rotation,  the 
three  lines,  or  parts,  KL,  KN,  KM,  as  radii,  will  de- 
scribe corresponding  circular  sections  of  these  solids, 
viz.,  KL  a  section  of  the  cylinder,  KN  a  section  of 
the  sphere,  and  KM  a  section  of  the  cone. 

Now,  FB  being  equal  to  FI  or  IG,  and  KM  paral- 
lel to  FB,  then,  by  similar  triangles,  IK  =  KM  (Geom. 
Theor.,  63),  and  IKN  is  a  right-angled  triangle ;  hence 
IN3  is  equal  to  IK2  +  KN2  (theor.  26).  But  KL  is 
equal  to  the  radius  IG  or  IN,  and  KM  =  IK  ;  there- 
fore KL2  is  equal  to  KM2  +  KN2,  or  the  square  of  the 
longest  radius  of  the  above-mentioned  circular  sec- 
tions is  equal  to  the  sum  of  the  squares  of  the  two 
others.  Now  circles  are  to  each  other  as  the  squares 
of  their  diameters,  or  of  their  radii,  therefore  the  cir- 
cle described  by  KL  is  equal  to  both  the  circles  de- 
scribed by  KM  and  KN ;  or  the  section  of  the  cylinder 
is  equal   to   both   the  corresponding  sections  of  the 


SOLID   GEOMETRY.  19 

sphere  and  cone.  And  as  this  is  always  the  case  in 
every  parallel  position  of  KL,  it  follows  that  the  cyl- 
inder EB,  which  is  composed  of  all  the  former  sec- 
tions, is  equal  to  the  hemisphere  EFG  and  cone  IAB, 
which  are  composed  of  all  the  latter  sections,  the  num- 
ber of  the  sections  being  the  same,  because  the  three 
solids  have  the  same  altitude. 

But  the  cone  IAB  is  a  third  part  of  the  cylinder  EB 
(Prop.  1 1,  cor.  3) ;  consequently,  the  hemisphere  EFG 
is  equal  to  the  remaining  two  thirds,  or  the  whole 
sphere  EFGH  is  equal  to  two  thirds  of  the  whole  cyl- 
inder ABCD. 

Corol.  1.  A  cone,  hemisphere,  and  cylinder  of  the 
same  base  and  altitude  are  to  each  other  as  the  num- 
bers 1,  2,  3.* 

Corol.  2.  All  spheres  are  to  each  other  as  the  cubes 
of  their  diameters,  all  these  being  like  parts  of  their 
circumscribing  cylinders. 

Corol.  3.  From  the  foregoing  demonstration  it  ap- 
pears that  the  spherical  zone  or  frustum  EGNP  is 
equal  to  the  difference  between  the  cylinder  EiGLO 
and  the  cone  IMQ,,  all  of  the  same  common  height 
IK.  And  that  the  spherical  segment  PFN  is  equal  to 
the  difference  between  the  cylinder  ABLO  and  the 
conic  frustum  AQMB,  all  of  the  same  common  alti- 
tude FK. 

Scholium.  By  the  scholium  to  Prop.  11,  we  have 
cone  A1B :  cone  QIM  : :  IF3 :  IK3 : :  FH3 :  (FH— 2FK)3 
.-.  cone  AIB  :  frust.  ABMQ : :  FH3 :  FH3— (FH— 2FK)3 
: :  fFH3 :  6FH\FK-  12FH.FKa+8FK3 ; 
but  cone  AIB  =  one  third  of  the  cylinder  ABGE ;  hence 
cvl.  AG :  frust.  ABMQ : :  3FH3 :  6FH\FK—  12FH.FK2 

+  8FK\ 
Nowcyl.AL:cyl.AG::   FK :    FI. 

Multiplying  the  last  two  proportions,  and  striking 
out  the  common  factors  from  the  ratios,  observing, 
also,  that  FI  =  ^FH,  we  have 

*  The  surfaces  of  the  sphere  and  circumscribing  cylinder  are  in  the 
ipme  ratio  as  their  solidities.  For  tlie  demonstration,  see  Mensuration, 
t  Raising  the  binomial  to  the  third  power. 


20  GEOMETRY. 

cyl.  AL  :  frust.  ABMQ : :  6FH2 :  6FH2— 12FH.FK  -f 

8FK2 
.*.  (dividendo,  and  by  corol.  3  of  this  Prop.,  14), 
cvl.  AL  :  segment  PFN  : :  6FH2: 12FH.FK— 8FKa 
::|FH2:FK(3FH— 2FK). 
But  cylinder  AL  =  circular  base  whose  diameter  is 
AB  or  FH,  multiplied  by  the  height  FK ;  hence  cyl- 
inder AL  =  circle  EFGH  X  FK. 

.•.segmentPFN=g.circI^fGH(3FH— 2FK)FK2. 
3         r  ri 


SPHERICAL  GEOMETRY. 


DEFINITIONS. 

1.  A  sphere  is  a  solid  terminated  by  a  curve  sur- 
face, and  is  such  that  all  the  points  of  the  surface  are 
equally  distant  from  an  interior  point,  which  is  called 
the  center  of  the  sphere. 

We  may  conceive  a  sphere  to  be  gen- 
erated by  the  revolution  of  a  semicircle 
APB  about  its  diameter  AB  ;  for  the  sur- 
face described  by  the  motion  of  the  curve 
APB  will  have  all  its  points  equally  dis- 
tant from  the  center  O. 

The  sector  of  a  circle  AOC  at  the  same 
time  generates  a  spherical  sector. 

2.  The  radius  of  a  sphere  is  a  straight 
line  drawn  from  the  center  to  any  point  on  the  surface. 

The  diameter  or  axis  of  a  sphere  is  a  straight  line 
drawn  through  the  center,  and  terminated  both  ways 
by  the  surface. 

It  appears  from  Def.  1  that  all  the  radii  of  the  same 
sphere  are  equal,  and  that  all  the  diameters  are  equal, 
and  each  double  of  the  radius. 

3.  It  will  be  demonstrated  (Prop.  1)  that  every 
section  of  a  sphere  made  by  a  plane  is  a  circle  ;  this 
being  assumed, 

A  great  circle  of  a  sphere  is  the  section  made  by  a 
plane  passing  through  the  center  of  the  sphere. 

A  small  circle  of  a  sphere  is  the  section  made  by  a 
plane  which  does  not  pass  through  the  center  of  the 
sphere. 

4.  The  pole  of  a  circle  of  a  sphere  is  a  point  on  the 
surface  of  the  sphere  equally  distant  from  all  the 
points  in  the  circumference  of  that  circle. 

It  will  be  seen  (Prop.  2)  that  all  circles,  whether 
jrreat  or  small,  have  two  poles. 


2  GEOMETRY. 

5.  A  spherical  triangle  is  the  portion  of  the  surface 
of  a  sphere  included  by  the  arcs  of  three  great  circles. 

6.  These  arcs  are  called  the  sides  of  the  triangle, 
and  each  is  supposed  to  be  less  than  half  of  the  cir- 
cumference. 

7.  The  angles  of  a  spherical  triangle  are  the  angles 
contained  between  the  planes  in  which  the  sides  lie. 
Or  the  angle  formed  by  any  two  arcs  of  great  circles 
is  the  angle  formed  by  the  planes  of  the  great  circles 
of  which  the  arcs  are  a  part. 

8.  A  spherical  polygon  is  the  portion  of  the  surface 
of  a  sphere  bounded  by  several  arcs  of  great  circles. 

9.  A  plane  is  said  to  be  a  tangent  to  a  sphere  when 
it  contains  only  one  point  in  common  with  the  surface 
of  the  sphere. 

10.  A  zone  is  the  portion  of  the  surface  and  a 
spherical  segment,  the  portion  of  the  volume  of  a 
sphere  between  two  parallel  planes,  or  cut  off'  by  one 
plane. 

The  circles  in  which  the  planes  intersect  the  sphere 
are  called  bases  of  the  zone  or  segment. 

11.  A  lune  is  the  portion  of  the  surface  of  a  sphere 
comprehended  between  two  great  semicircles. 

12.  A  spherical  wedge  or  ungulais  the  solid  bound- 
ed by  a  lune  and  the  planes  of  its  two  circles. 

PROP.   I. 

Every  section  of  a  sphere  made  by  a  plane  is  a  cir- 
cle. 

Let  AZBX  be  a  sphere  whose  A 

center  is  O. 

Let  XPZ  be  a  section  made  by 
the  plane  XZ.  x; 

From  O  draw  OC  perpendicular 
to  the  plane  XZ. 

In  XPZ  take  any  points  Px,  P2, 

P3 

JoinCP,;  CP2:  CP3; 

also,  OP, ;  OP2 ;  OP3 ; 


SPHERICAL    GEOMETRY.  J 

Then,  since  OC  is  perpendicular  to  the  plane  XZ, 
it  will  be  perpendicular  to  all  straight  lines  passing 
through  its  loot  in  that  plane.  (Dei.  3,  Geometry  of 
Planes.) 

Hence  the   angles   OCPi,   OCP2,  OCP3 

are  right  angles 

OP^CP^  +  OC2; 
OP22  =  CP23  +  OC2; 
OP32  =  CP33  +  OC\ 

But,  since  PM  P2,  P3 are  all  points  upon 

the  surface  of  the  sphere,  v  by  def.  1,  OP1  =  OP2  = 

OP3= 

...  CP1  =  CP2  =  CP3 

Hence  XPZ  is  a  circle  whose  center  is  C,  and  every 
other  section  of  a  sphere  made  by  a  plane  may,  in 
like  manner,  be  proved  to  be  a  circle. 

Cor.  1.  If  the  plane  pass  through  the  center  of  the 
sphere,  then  OC  =  0,  and  the  radius  of  the  circle  will 
be  equal  to  the  radius  of  the  sphere. 

Cor.  2.  Hence  all  great  circles  are  equal  to  one 
another,  since  the  radius  of  each  is  equal  to  the  ra- 
dius of  the  sphere. 

Cor.  3.  Hence,  also,  two  great  circles  and  their  cir- 
cumferences always  bisect  each  other  ;  for,  since  both 
pass  through  the  center,  their  common  intersection 
passes  through  the  center,  and  is  a  diameter  of  the 
sphere  and  of  each  of  the  two  circles. 

Cor.  4.  The  center  of  a  small  circle  and  that  of  the 
sphere  are  in  a  straight  line,  which  is  perpendicular 
to  the  plane  of  the  small  circle. 

Cor.  5.  We  can  always  draw  one,  and  only  one, 
great  circle  through  any  two  points  on  the  surface  of 
a  sphere ;  for  the  two  given  points  and  the  center  of 
the  sphere  give  three  points,  which  determine  the  po- 
sition of  a  plane. 

If,  however,  the  two  given  points  are  the  extremi- 
ties of  a  diameter,  then  these  two  points  and  the  cen- 
ter of  the  sphere  are  in  the  same  straight  line,  and  an 
infinite  number  of  great  circles  may  be  drawn  through 
the  two  points.     (Prop.  3,  Geom.  of  Planes.) 


GEOMETRY 


Distances  on  the  surface  of  a  sphere  are  measured 
by  the  arcs  of  great  circles.  The  reason  for  this  is, 
that  the  shortest  line  which  can  be  drawn  upon  the 
surface  of  a  sphere,  between  any  two  points,  is  the 
arc  of  a  great  circle  joining  them,  which  will  be 
proved  hereafter. 

PROP.   II. 

If  a  diameter  be  drawn  perpendicular  to  the  plane 
oj  a  great  circle,  the  extremities  of  the  diameter  will  be 
the  poles  of  that  circle,  and  of  all  the  small  circles  whose 
planes  are  parallel  to  it. 

Let  APB  be  a  great  circle  of  z 

the  sphere  whose  center  is  O. 

Draw  ZN,  a  diameter  perpen- 
dicular to  the  plane  of  the  circle 
APB. 

Then  Z  and  N,  the  extremities 
of  this  diameter,  are  the  poles  of 
the  great  circle  APB,  and  of  all 
the  small  circles,  such  as  apb, 
whose  planes  are  parallel  to  that  of  APB. 

Take  any  points  P„  P2, in  the  circumfe- 
rence of  APB, 

Through  each  of  these  points  respectively,  and  the 
points  Z  and  N,  describe  great  circles,  ZP,N,  ZP.N. 

Join  OP,,  OP2, 

Then,  since  ZO  is  perpendicular  to  the  plane  of 
APB,  it  is  perpendicular  to  all  the  straight  lines  OP^ 
OP.2, drawn  through  its  foot  in  that  plane. 

Hence  all  the  angles  ZOP„  ZOP,, are 

right  angles,  and  .-.  the  arcs  ZPX,  ZP2, are 

quadrants. 

Thus  it  appears  that  the  points  Z  and  N  are  at  a 
quadrant's  distance,  and  .*.  equally  distant  from  all 
the  points  in  the  circumference  of  APB,  and  are  .*. 
the  poles  of  that  great  circle. 

Again;  since  ZO  is  perpendicular  to  the  plane  APB, 
it  is  also  perpendicular  to  the  parallel  plane  apb  (Ge- 
ometry of  Planes,  Prop.  14). 


sp;ii:r[cal  geometry.  o 

Hence  the  oblique  lines  Zp{,  7,p:, drawn 

to  pt,  p:,  in  the  circumference  of  apb,  will  be  equal  to 
each  other.     (Prop.  7,  Geometry  of  Planes.) 

.*.  The  chords  Zpn  Zp„ being  equal,  the 

arcs  Zp„  Zp2, which  they  subtend,  will  also 

be  equal. 

.-.  The  point  Z  is  the  pole  of  the  circle  apb;  and 
the  point  N  is  also  a  pole,  the  arcs  Np„  &c,  being  sup- 
plements of  the  arcs  Z/?„  &c. 

Def.  The  diameter  of  the  sphere  perpendicular  to 
the  plane  of  a  circle  is  called  the  axis  of  that  circle. 

Cor.  1.  Every  arc  PiZ,  drawn  from  a  point  in  the 
circumference  of  a  great  circle  to  its  pole,  is  a  quad- 
rant, and  this  arc  PiZ  makes  a  right  angle  with  the 
arc  AP,B.  For,  the  straight  line  ZO  being  perpen- 
dicular to  the  plane  APB,  every  plane  which  passes 
through  this  straight  line  will  be  perpendicular  to  the 
plane  APB  (Prop.  18,  Geometry  of  Planes)  ;  hence 
the  angle  between  these  planes  is  a  right  angle,  or,  by 
def.  7,  the  angle  of  the  arcs  APi  and  ZPj  is  a  right 
angle. 

Cor.  2.  In  order  to  find  the  pole  of  a  given  arc  APX 
of  a  great  circle,  take  PiZ,  perpendicular  to  A  Pi,*  and 
equal  to  a  quadrant,  the  point  Z  will  be  a  pole  of  the 
arc  AP, ;  or,  from  the  points  A  and  Pt  draw  two  arcs 
AZ  and  P,Z  perpendicular  to  AP1?  the  point  Z  in  which 
they  meet  is  a  pole  of  APj. 

Cor.  3.  Reciprocally,  if  the  distance  of  the  point 
Z  from  each  of  the  points  A  and  Px  is  equal  to  a  quad- 
rant, then  the  point  Z  is  the  pole  of  APi,  and  each  of 
the  angles  ZAPM  ZPiA  is  a  right  angle. 

For,  let  O  be  the  center  of  the  sphere ;  draw  the 
radii  OA,OP„OZ; 

Then,  since  the  angles  AOZ,  PrOZ  are  right  angles, 
the  straight  line  OZ  is  perpendicular  to  the  straight 
lines  OA,  OPi,  and  is  .*.  perpendicular  to  their  plane  ; 
hence,  by  the  above  prop.,  the  point  Z  is  the  pole  of 

*  A  perpendicular  arc  to  APi  at  Pi  is  described  by  means  of  its 
pole,  which  will  be  in  Al'iB,  at  a  quadrant's  distance  From  Pi. 


6  GEOMETRY. 

AP1?  nnd  .*.  (corol.  1),  the  angles  ZAP15  ZPXA  are  right 
angles. 

Cor.  4.  Great  circles,  such  as  ZA,  ZPj,  whose 
planes  are  at  right  angles  to  the  plane  of  another 
great  circle,  as  APB,  are  called  its  secondaries  ;  and 
it  appears  from  the  foregoing  corollaries,  that, 

1.  The  planes  of  all  secondaries  pass  through  the 
axis,  and  their  circumferences  through  the  poles  of 
the^r  primary ;  and  that  the  poles  of  any  great  circle 
may  always  be  determined  by  the  intersection  of  any 
two  of  its  secondaries. 

2.  The  arcs  of  all  secondaries  intercepted  between 
the  primary  and  its  poles  are  =90°. 

3.  A  secondary  bisects  all  circles  parallel  to  its  pri- 
mary, the  axis  of  the  latter  passing  through  all  their 
centers. 

Cor.  5.*  Let  the  radius  of  the  sphere  =  R,  radius 
of  small  circle  parallel  to  it  =  r.  Distance  of  two  cir- 
cles, or  Oo  =  6. 

Join  Opl9  and  let  the  arc  P,^,  in  degrees  and  frac- 
tions of  a  degree,  be  expressed  by  0.  Then  will  6 
=  sin.  0  and  r  =  cos.  0  to  the  radius  R,  and  we  have 
the  equations 

R*  =  r"  +   <f  ; 
r    =  R  cos.  0  ; 
d    =  R  sin.  0  ; 
in  which  cos.  0  and  sin.  0  express  these  trigonomet- 
rical lines  to  radius  1;  the  usual  radius  of  the  tables. 

Cor.  6.  Two  secondaries  intercept  similar  arcs  of 
circles  parallel  to  their  primary,  and  these  arcs  are 
to  each  other  as  the  cosines  of  the  arcs  of  the  sec- 
ondaries between  the  parallels  and  the  primary. 

For  the  arcs  of  the  parallels  subtend  at  their  re- 
spective centers,  angles  equal  to  the  inclinations  of 
the  planes  of  the  secondaries,  and  these  arcs  will, 
therefore  (def.  55),  be  similar.  Again  :  let  p{p2  in  the 
diagram  be  one  of  these  arcs,  and  imagine  another, 


*  The  two  following  corollaries  require  a  knowledge  of  the  first 
principles  of  Trigonometry. 


SPHERICAL    GEOMETRY. 


fto*  between  this  and  PiP, ;  then  if  r„  r3  be  the  radii 
of1  the  two  small  parallels  ptp„  q{q2,  the  rest  of  nota- 
tion as  before,  we  shall  have 

arc  ptfi     whole  circumference  of  1st  # 
arc  q^,  ~  whole  circumference  of  2d 

=  -(th.  71,  cor.  1); 

_  R  cos.  <j> 

R  cos.  <f>' ' 
_  cos.  <p 

~~  COS.  0'" 

If  the  second  arc  qvq3  becomes  PjP2,  <f>f  =  0  and  cos. 

<f>f=l. 

arc  P.P.         1  arc  »,»2  , 

Cfty-  =  cos.  0,  or  arcp,p3  = 

■TlM 


arc  p{p 
cos.  0  arc  P^a.* 


r»  or 

cos.  <p        arc 


PROP.   III. 

Every  plane  perpendicular  to  a  radius  at  its  extrem- 
ity is  a  tangent  to  the  sphere  in  that  point. 

Let  ZXY  be  a  plane  perpendicular  to 
the  radius  OZ. 

Then  ZXY  touches  the  sphere  in  Z. 

Take  any  point  P  in  the  plane ;  join 
ZP  •  OP  • 

Then  (Prop.  6,  Geom.  of  Planes)  OP 
>OZ. 

Hence   the  point  P  is  without   the 
sphere  ;  and,  in  like  manner,  it  may  be 
shown  that  every  point  in  XYZ,  except  Z,  is  without 
the  sphere. 

Therefore  the  plane  XYZ  is  a  tangent  to  the 
sphere. 

*  These  formulas  are  of  frequent  use  in  Astronomy,  serving  to  ex- 
press the  relation  between  the  distance  moved  on  a  parallel  of  decli- 
nation and  in  right  ascension  of  a  star,  and  various  other  useful  relar 
tions  of  a  similar  kind. 


GEOMETRY. 


ntor.  IV. 


The  angle  formed  by  two  arcs  of  great  circles  is 
equal  to  the  angle  contained  by  the  tangents  drawn  to 
these  arcs  at  their  point  of  intersection,  and  is  measured 
by  the  arc  described  from  their  point  of  intersection  as 
a  pole,  and  intercepted  between  the  arcs  containing  the 


angle. 


Let  ZPN,  ZQN,  arcs  of  great 
circles,  intersect  in  Z. 

Draw  ZT,  ZT',  tangents  to  the 
arcs  at  the  point  Z. 

With  Z   as   pole,  describe  the 
arc  PQ. 

Take  O,  the  center  of  the  sphere, 
and  join  OP,  OQ. 

Then  the  spherical  angle  PZQ 
is  equal  to  the  angle  TZT',  and  is  measured  by  the 
arc  PQ. 

For  the  tangent  ZT,  drawn  in  the  plane  ZPN,  is 
perpendicular  to  radius  OZ  ;  and  the  tangent  ZT', 
drawn  in  the  plane  ZQN,  is  perpendicular  to  radius 
OZ  ;  hence  the  angle  TZT'  is  equal  to  the  angle  con- 
tained by  these  two  planes  (def.  6,  Geom.  of  Planes), 
that  is,  to  the  spherical  angle  PZQ. 

Again ;  since  the  arcs  ZP,  ZQ  are  each  of  them 
equal  to  a  quadrant ; 

.'.  Each  of  the  angles  ZOP,  ZOQ  is  a  right  angle, 
or  OP  and  OQ  are  perpendicular  to  ZO. 

.-.  The  angle  QOP  is  the  angle  contained  by  the 
planes  ZPN,  ZQN. 

.*.  The  arc  PQ,  which  measures  the  angle  POQ, 
measures  the  angle  between  the  planes,  that  is,  the 
spherical  angle  PZQ. 

Cor.  1.  The  angle  under  two  great  circles  is  meas- 
ured by  the  distance  between  their  poles.  For  the 
axes  (def.  in  Prop.  2)  of  the  great  circles  drawn 
through  their  poles  being  perpendicular  to  the  planes 
of  the  circles,  will  be  perpendicular  to  all  lines  of 


SPHERICAL   GEOMETRY. 


these  planes,  consequently,  to  the  lines  which  measure 
the  angles  of  the  planes,  and  .*.  (see  th.  65,  Gen.  Sch., 
4°)  the  angles  under  these  axes  will  be  equal  to  the 
angle  between  the  circles  ;  but  the  angle  under  the 
axes  is  obviously  measured  by  the  arc  which  joins 
their  extremities,  that  is,  by  the  distance  between 
their  poles. 

Cor.  2.  The  angle  under  two  great  circles  is  meas- 
ured by  the  arc  of  a  common  secondary  intercepted 
between  them.  Q 

Cor.  3.  Vertical  spherical  an- 
gles, such  as  QP  W,  RPS,  are 
equal,  for  each  of  them  is  the 
angle  formed  by  the  planes  QP 
R,  WPS. 

Also,  when  two  arcs  cut  each  w 
other,  the  two  adjacent  angles 
QPW,  QPS,  when   taken  to- 
gether, are  always  equal  to  two 
right  angles. 


prop.  v. 

If  from  the  angular  points  of  a  spherical  triangle  con- 
sidered as  poles,  three  arcs  be  described  forming  another 
triangle,  then,  reciprocally,  the  angular  points  of  this 
last  triangle  will  be  the  poles  of  the  sides  opposite  to 
them  in  the  first. 

Let  ABC  be  a  spherical 
triangle. 

From  the  points  A,  B,  C, 
considered  as  poles,  describe 
the  arcs  EF,  DF,  DE,  form- 
ing the  spherical  triangle  D 
EF. 

Then  D  will  be  the  pole  of 
BC,  Eof  AC,  andFofAB. 

For,  since  B  is  the  pole  of 
DF,  the  distance  from  B  to  D 
is  a  quadrant. 

TT 


10 


GEOMETRY. 


And,  since  C  is  the  pole  of  DE,  the  distance  from 
C  to  D  is  a  quadrant. 

Thus,  it  appears  that  the  point  D  is  distant  by  a 
quadrant  from  the  points  B  and  C. 

.-.  (cor.  1,  2,  Prop.  2)  D  is  the  pole  of  the  arc  BC. 

Similarly,  it  may  be  shown  that  E  is  the  pole  of 
AC,  and  F  the  pole  of  AB. 

Note.  D  having  been  shown  to  be  the  pole  of  the 
arc  passing  through  the  points  B  and  C,  it  must  be  of 
the  arc  BC,  because  but  one  arc  of  a  great  circle  can 
be  made  to  pass  through  the  two  points  B  and  C  (cor. 
5,  Prop.  1). 

PRor.  vi. 

The  same  things  being  given  as  in  the  last  proposi- 
tion, each  angle  in  either  of  the  triangles  will  be  meas- 
ured by  the  supplement  of  the  side  opposite  to  it  in  the 
other  triangle. 

Produce  BC  to  I  and  K,  AB 
to  G,  and  AC  to  H. 

Then,  since  A  is  the  pole 
of  EF,  the  angle  A  is  measur- 
ed by  the  arc  GH  at  a  quad- 
rant's distance  from  A  (Prop. 

4). 

But,  because  F  is  the  pole 
of  AG,  the  arc  FG  is  a  quad- 
rant. 

And,  because  E  is  the  pole 
of  AH,  the  arc  EH  is  a  quadrant. 
.-.  EH  +  GF=180°, 
or  EF+GH  =  180°; 

.-.  GH  =  180°  —  EF. 

In  a  similar  manner,  it  may  be  proved  that  the  an- 
gle B  is  measured  by  180°  —  DF,  and  the  angle  C  by 
180°  —  DE. 

Again;  since  D  is  the  pole  of  BC,  the  angle  D  is 
measured  by  IK. 

But,  because  B  is  the  pole  of  DK,  the  arc  BK  is  a 
quadrant. 


SPHERICAL    GEOMETRY.  11 

And,  because  C  is  the  pole  of  DI,  the  arc  CI  is  a 
quadrant. 

.-.  IC  +  BK=180°, 
or  IK+BC  =  180°; 

.-.  IK  =180°  —  BC. 

But  IK  is  the  measure  of  the  angle  D  (Prop.  4). 

In  the  same  manner,  it  may  be  proved  that  the  an- 
gle E  is  measured  by  180°  — AC,  and  the  angle  F  by 
180°  — AB. 

These  triangles  ABC,  DEF  are,  from  their  prop- 
erties, usually  called  Polar  triangles,  or  Supplemental 
triangles. 

PROP.    VII. 

In  any  spherical  triangle  any  one  side  is  less  than 
the  sum  of  the  other  two. 

Let  ABC  be  a  sphercal  triangle,        A 
O  the  center  of  the  sphere.     Draw 
the  radii  OA,  OB,  OC. 

Then  the  three  plane  angles  AOB, 
AOC,  BOC  form  a  trihedral  angle  at 
the  point  O,  and  these  three  angles 
are  measured  by  the  arcs  AB,  AC,  BC. 

But  each  of  the  plane  angles  which  6 
form  the  trihedral  angle  is  less  than  the  sum  of  the 
two  others  (Prop.  1,  Polyhedral  Angles). 

Hence  each  of  the  arcs  AB,  AC,  BC,  which  meas- 
ures these  angles,  is  less  than  the  sum  of  the  other 
two. 

PROP.    VIII. 

The  sum  of  the  three  sides  of  a  spherical  triangle  is 
less  than  the  circumference  of  a  great  circle. 

Let  ABC  be  any  spherical  tri- 
angle. 

Produce  the   sides  AB,  AC   to 
meet  in  D. 

Then,  since  two  great  circles  al- 
ways bisect  each  other  (Prop.  \$£< 
cor.  3),  the  arcs  ABD,  ACD  are 
semicircles. 


12  GEOMETRY. 

Now,  in  the  triangle  BCD, 

BC  <  BD  +  DC,  by  last  Prop. 
...  AB  +  AC  +  BC  <  AB  +  BD  +  AC  +  CD, 

<  ABD  +  ACD, 

<  circumference  of  great  circle. 
Note.  In  elementary  geometry  the  only  spherical 

triangles  considered  are  those  in  which  each  side  is 
less  than  a  semicircumference,  and  each  angle  less 
than  two  right  angle's. 

Should  a  spherical  triangle  be  taken  without  these 
restrictions,  it  will  be  found  that  the  residue  of  the 
surface  of  the  sphere  will  be  a  triangle  having  portions 
of  the  same  circumferences  as  boundaries  with  the 
given  triangle,  and  falling  within  the  restrictions ; 
when  all  the  parts  of  this  latter  triangle  are  known, 
the  parts  of  the  other  may  be  derived  from  them  by 
subtracting  the  known  angles  and  the  known  sides 
from  180  or  360  degrees. 

Triangles  not  limited  by  the  restrictions  above 
mentioned,  therefore,  being  dependent  upon  those 
which  are  thus  limited,  the  first  class  may  be  reject- 
ed, and  our  attention,  as  it  has  been  in  the  preceding 
theorems,  confined  to  the  second. 

PROP.    IX. 

Two  spherical  triangles  are  either  identical  or  sym- 
metrical, 1°.  When  they  have  two  sides  and  the  includ- 
ed angle  of  the  one  equal  to  the  same  in  the  other  ;  2°. 
When  they  have  a  side  and  two  adjacent  angles  ;  3°. 
When  they  have  three  sides  respectively  equal. 

These  follow  from  the  cor-  A 
responding  theorems  in  trihe- 
dral angles  (Prop.  3,  and  exer. 
3,  4,  5),  but  may  be  proved  by 
superposition  of  the  given  tri- 
angles, the  one  upon  the  other, 
or  its  symmetrical  triangle,  as 
at  th.  1,  2,  &c,  in  Plane  Geom- 
etry. 

The  preceding  diagram  exhibits  symmetrical  tri- 


SPHERICAL  GEOMETRY. 


13 


Ml- 


angles,  viz.,  EDF  and  EDF',  or  ABC  and  EDF.  The 
one  could  not  be  superposed  upon  the  other,  for,  on 
turning  it  over  so  as  to  bring  the  equal  parts  opposite 
to  each  other,  the  convexities  of  the  two  surfaces 
would  be  turned  toward  each  other,  and  could  touch 
in  but  one  point. 

prop.  x. 

Symmetrical  triangles  are  nevertheless  equal  in 
surface,  which  may  be  proved  as  follows : 

Let  ABC,  DEF  be  two  A  D 

symmetrical  triangles,  in 
which  AB  =  DE,  AC  =  DF, 
BC  -  EF. 

Let  G  be  the  pole  of  the 
small  circle  passing  through  B^ 
the  three  points  A,  B,  C,  and 
H  the  pole  of  the  small 
circle  passing  through  the 
three  D,  E,  F.  Join  G  with  A,  B,  C,  and  H  with  D, 
E,  F  by  arcs  of  great  circles.  The  triangles  AGB, 
BGC,  AGC,  HDE,  HFE,  HDF  are  all  isosceles,  and 
the  corresponding  ones  in  the  two  diagrams  admit  of 
superposition,  because,  in  turning  them  over  to  bring 
the  convexities  of  their  surfaces  the  same  way,  equal 
sides  are  not  turned  away  from  each  other,  and  this 
arises  from  the  triangles  being  isosceles.  The  three 
triangles  of  the  one  diagram  being  respectively  iden- 
tical, therefore,  with  the  three  of  the  other,  we  have 
AGB  +  BGC  —  AGC  =  DHE  +  EHF  —  DHF,  or  A 
BC  =  DEF.    Q.  E.  D. 

PROP.   XI. 

An  isosceles  spherical  triangle  has  its  two  angles 
equal,  and  conversely. 

PROP.    XII. 

In  any  spherical  triangle,  the  greater  side  is  opposite 
the  greater  angle,  and  conversely. 

These  may  be  proved  precisely  as  in  plane  trian- 
gles. 


14  GEOMETRY. 


PROP.    XIII. 


Two  spherical  triangles  (unlike  two  plane  triangles 
in  this  respect)  are  equal  when  the  three  angles  of  the 
one  are  equal  to  the  three  angles  of  the  other,  each  to 
each. 

For  the  polar  triangles  of  the  two  given  triangles 
will  have  equal  sides  (Prop.  6),  and,  consequently, 
equal  angles  (Prop.  9).  Hence  the  given  triangles 
will  have  equal  sides. 

NOTE. 

The  equal  triangles  in  question  in  the  preceding 
theorems  need  not  be  supposed  on  the  same  sphere,  if 
their  sides  and  angles  are  given  in  degrees  and  frac- 
tions of  a  degree.  Indeed,  there  would  be  much 
advantage  gained  by  discarding  spherical  triangles 
from  geometry  except  for  purposes  of  mensuration  on 
the  surface  of  the  sphere,  and  using  trihedral  angles 
in  their  place,  especially  in  the  application  to  Astron- 
omy, which,  as  a  science  of  observation,  depends  en- 
tirely on  angular  measurements. 

PROP.   XIV. 

The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  two  and  less  than  six  right  angles. 

For  each  angle  is  less  than  two  (note  to  Prop.  8), 
hence  the  sum  of  the  three  is  less  than  six  right  an- 
gles. Again,  each  angle  being  measured  by  a  semi- 
circumference,  minus  the  side  opposite  in  the  polar 
triangle,  the  sum  of  the  three  angles  will  be  three 
semicircumferences,  minus  the  sum  of  the  three  sides 
of  the  polar  triangle,  but  the  latter  is  less  than  a  cir- 
cumference (Prop.  8) ;  hence  the  measure  of  the  sum 
of  the  three  angles  will  be  greater  than  one  semicir- 
cumference  or  two  right  angles. 

Note.  In  a  birectangular  spherical  triangle,  two  of 
the  sides  are  quadrants ;  and  in  a  trirectangular  tri- 


SPHERICAL   GEOMETRY.  15 

angle  all  three  of  the  sides  are  quadrants.  This  latter 
triangle  is  sometimes  taken  as  the  unit  of  measure  on 
the  surface  of  the  sphere.  As  there  are  four  such 
triangles  in  each  hemisphere,  the  whole  surface  of  the 
sphere  would  be  expressed  by  the  number  8. 

trop.  xv. 

The  surface  of  a  lune  is  to  the  whole  surface  of  the 
sphere  as  the  angle  of  the  lune  is  to  four  right  angles, 
or  as  the  arc  which  measures  the  angle  of  the  lune  is  to 
a  circumference. 

It  is  evident,  from  a  mere  in- 
spection of  the  diagram,  thatthe 
lune  ABDC  is  the  same  aliquot 
part  of  the  whole  surface  of 
the  sphere  that  the  arc  BC  is  of 
a  whole  circumference,  or  that 
the  angle  BAG,  measured  by 
this  arc,  is  of  four  right  angles. 
The  demonstration  may  be 
made  more  full  by  dividing  the 
triangle  ABC  into  a  number 
of  equal  triangles,  having  their  common  vertex  at  A 
and  their  bases  equal  portions  of  BC,*  and  dividing, 
also,  the  hemisphere  into  triangles  of  the  same  size, 
and  thus  showing  that  the  ratio  of  the  triangle  ABC 
to  the  hemisphere  is  the  same  as  the  ratio  of  BC  to 
the  whole  circumference,  because  both  are  in  the 
ratio  of  the  same  two  numbers,  viz.,  the  number  of 
triangles  in  ABC  to  the  number  in  the  hemisphere,  or 
the  number  of  bases  in  each. 

Schol.  The  angle  of  the  lune  is  to  four  as  twice  this 
angle  is  to  eight.  Hence,  if  the  whole  sphere  be  ex- 
pressed by  8,  the  lune  will  be  expressed  by  2A. 

*  These  triangles  will  be  equal  because  their  sides  are  equal. 


16 


GEOMETRY. 


PROP.   XVI. 

The  two  opposite  spherical  triangles  on  a  hemisphere 
are  together  equal  to  a  lune  having  the  same  angle. 

Let  the  two  triangles  ABC,  ADE 
be  on  the  same  hemisphere,  having 
their  common  vertex  at  A.  Then 
will  their  sum  be  equal  to  the  lune 
ABFC. 

For  the  triangle  BFC  may  be 
proved  equilateral  with  the  trian- 
gle ADE,  and,  therefore,  of  the 
same  surface.    Q.  E.  D. 


PROP.   XVII. 

The  measure  of  a  spherical  triangle  is  the  excess  of 
the  sum  of  its  angles  above  two  right  angles. 

Let  ABC  be  a  spherical  trian- 
gle ;  its  measure  will  be  A  +  B  + 
C— 2.* 

For  (by  the  last  two  proposi- 
tions), 

ADAE+  AGAH  =  2A; 
AFBG  +  AlBD   =2B; 
aHCI  +  aECF  =2C. 
If  we  add  the  first  members  to- 
gether, we  obtain  evidently  the  whole  hemisphere, 
which  is  expressed  by  four,  together  with  twice  the 
triangle  ABC. 

.-.  4  +  2  A  ABC  =  2 A  +  2B  +  2C  ; 

ABC=   A  +    B+    C  — 2. 

Q.  E.  D. 
Corol.  1.  The  spherical  triangle  is  equivalent  to  a 
lune  whose  angle  is  half  the  above  expression. 

Corol.  2.  Two  spherical  triangles  are  of  equal  sur- 
face when  the  sum  of  their  angles  is  the  same,  and 
vice  versct. 

*  A,  B,  and  C  must  be  here  understood  as  expressed  not  in  de- 
grees, &c.,  but  in  fractions  of  a  right  angle. 


EXERCISES.  17 


EXERCISES. 

1.  Prove  that  every  spherical  triangle  may  be  inscribed  in  a  circle. 

2.  Through  a  given  point  on  the  arc  of  a  great  circle  to  draw  an 
arc  of  a  great  circle  perpendicular  to  the  former. 

3.  The  same  through  a  point  without  the  given  arc. 

4.  Prove  that  the  rectangles  of  the  parts  of  all  lines  passing  through 
the  same  point  within  a  sphere,  and  terminating  at  the  surface,  are 
equal. 

5.  Trove  that  circles  whose  planes  are  equidistant  from  the  center 
of  the  sphere  are  equal. 

6.  Prove  that  every  plane  passing  through  the  point  of  contact  of  a 
tangent  piano  to  a  sphere  cuts  this  plane  in  a  line  tangent  to  the  circle 
cut  from  the  sphere. 

7.  That  the  line  of  centers  of  two  spheres  which  cut  each  other  is 
perpendicular  to  the  plane  of  the  circle  of  section  of  the  two  spheres. 

8.  Prove  that  their  intersection  is  a  circle. 

9.  Show  how  to  construct  a  spherical  triangle  with  any  three  parts 
given. 

10.  Prove  that  the  sum  of  all  the  sides  of  a  spherical  polygon  is  less 
than  the  circumference  of  a  great  circle. 

11.  Make  a  sphere  pass  through  four  given  points,  or  prove  that 
every  tetrahedron  may  be  circumscribed  by  a  sphere. 

12.  Also,  inscribed. 

13.  Prove  that  the  measure  of  the  surface  of  a  spherical  polygon  is 
equal  to  the  excess  of  the  sum  of  its  angles  over  as  many  times  two 
right  angles  as  the  Bgure  has  sides  less  two. 

14.  Make  a  great  circle  tangent*  to  a  small  circle  on  the  surface 
of  a  sphere. 

15.  Change  a  spherical  quadrangle  into  an  equivalent  spherical 
triangle. 

16.  Upon  the  base  of  a  spherical  triangle  to  construct  an  isosceles 
spherical  triangle  of  equal  surface. 

17.  To  construct  on  the  base  of  a  given  spherical  triangle  another 
of  equal  surface,  1°,  having  a  given  base  angle;  2°,  having  a  given 
side. 

18.  Prove  that  the  sums  of  the  opposite  angles  of  a  spherical  quad- 
rilateral inscribed  in  a  circle  of  the  sphere  are  equal.t 

*  One  circle  is  said  to  be  tangent  to  another  on  the  surface  of  a 
sphere  when  the  two  circles  have  a  common  tangent  line  at  a  com- 
mon point. 

t  This  is  done  by  connecting  the  four  vertices  of  the  quadrilateral 


18  GEOMETRY. 

19.  Prove  that  if  two  spherical  triangles,  having  a  common  base,  be 
inscribed  in  the  same  circle  of  a  sphere,  the  difference  between  the 
sum  of  the  base  angles  and  the  vertical  angle  will  be  equal  in  the 
two  triangles. 

Corol.  Spherical  triangles  having  the  same  base,  and  the  sums  of 
their  base  angles  equal,  and  also  their  vertical  angles  equal,  have  their 
vertices  lying  in  the  same  circumference  on  the  sphere. 

20.  Prove  that  if  the  base  of  a  spherical  triangle  be  prolonged  to 
become  a  complete  circumference,  and  the  other  two  sides  prolonged 
beyond  the  vertex  till  they  meet  this ;  then,  if  through  the  points  of 
meeting  and  the  vertex  a  small  circle  of  the  sphere  be  made  to  pass, 
every  triangle  having  its  vertex  in  this,  and  its  base  the  same  with 
the  given  triangle,  will  have  an  equal  surface. 

Corol.  If  one  of  the  other  sides  of  the  triangle  falls  in  the  prolonga- 
tion of  the  base,  and  the  vertex  coincides  with  one  of  the  above- 
mentioned  points  of  meeting,  the  small  circle  passing  through  the 
three  points  vanishes  or  reduces  to  a  point,  viz.,  the  point  in  which 
these  three  points  coalesce  ;  the  triangle  then  degenerates  into  a  lune, 
which  is  still,  however,  equal  to  the  given  triangle  in  surface. 

21.  Upon  the  base  of  a  given  spherical  triangle  to  construct  another 
of  equal  surface  of  which  the  vertex  shall  he  in  a  given  great  circum- 
ference. 

22.  To  change  a  spherical  triangle  into  another  of  equal  surface 
with  a  given  side  and  given  angle  adjacent. 

23.  To  construct  a  spherical  triangle  with  two  given  sides  and  of 
surface  equal  to  a  given  triangle. 

24.  Prove  that  if  P  denote  the  number  of  polyhedral  angles  of  a 
polyhedron,  F  the  number  of  its  faces,  and  E  the  number  of  its  edges, 

P  +  F  =  E-f  2. 

25.  Also,  that  the  sum  of  the  plane  angles  of  a  polyhedron  is  equal 
to  P  —  2  times  four  right  angles. 

26.  To  construct  the  length  of  the  radius  of  a  sphere  when  con- 
fined to  its  exterior. 

27.  To  describe  the  circumference  of  a  great  circle  through  two 
given  points. 

with  the  pole  of  the  circle  in  which  it  is  inscribed,  thus  forming  four 
isosceles  spherical  triangles. 


APPENDIX  III. 


Prove  that  if  _  expresses  the  ratio  of  an  arc  to  a  quadrant, 

n  n 


m   n 
~"2 


will  express  the  ratio  of  the  arc  to  the  radius. 

Knowing  the  ratio  of  an  arc  to  the  radius,  show  how  to  find  the 


Two  angles  subtended  by  arcs  of  different  radii  are  to  each  other  as 
the  ratios  of  the  arcs  to  their  respective  radii  (th.  71,  corol.  3).  In 
symbols,  if  V  and  V  be  two  angles,  A  and  A'  the  arcs  subtending 
them,  described  with  the  radii  R  and  R', 

V*  V  •  •  —  •      . 

"  R '  R' 

Taking  the  right  angle  as  the  unit  of  angles,  supposing  for  a  moment 
V  to  be  this  unit,  A'  the  unit  of  arc,  and  R'  the  unit  of  length,  the 
above  proportion  becomes 

V:l::^:l  .-.  V  =  ^; 
R  R' 

that  is,  an  angle  at  the  center  has  for  its  measure  the  quotient  of  the 
arc  which  subtends  it,  divided  by  the  radius.  It  must,  however,  be 
understood  that  the  quantities  V,  A,  and  R  are  referred  to  their  re- 
spective units. 


Of  two  arcs,  each  less  than  a  semicircumferen.ee,  subtended  by  the 
same  chord,  the  shortest  is  that  whose  center  is  furthest  from  the  middle 
of  the  chord. 

Let  AB  be  the  common  chord,  AMB, 
AM'B  the  two  arcs,  O  the  center  of  the  for- 
mer, O'  of  the  latter.  Then,  if  OP  >  O'P, 
AMB  <  AM'B. 

For  (by  th.  17)  OA>0'A;  and,  if  the 
arc  AM'B  be  turned  over  round  AB  as  a 
hinge,  it  will  evidently  contain  the  arc 
AMB   within  it;*   and  it  may  be  easily 

*  This  may  be  seen  more  distinctly  by 
observing  that  an  indefinitely  small  portion 
of  the  arc  of  a  circle  may  be  regarded  as  a 


2  GEOMETRY. 

proved,  that  of  two  lines,  the  one  enveloping  the  other,  and  termina- 
ting at  the  same  points,  the  enveloped  line  is  the  least.  This  maybe 
shown,  supposing  them  to  be  polygonal  lines  at  first,  by  repeated  ap- 
plication of  the  principle  that  a  straight  line  is  the  shortest  distance 
between  two  points,  and  then  supposing  the  straight  portions  of  the 
polygonal  lines  to  become  infinitely  small,  or  the  polygonal  lines  to 
become  curves. 

Prove  that  every  small  circle  of  a  sphere  has  a  less  radius  than  the 
sphere. 

THEOREM. 

The  arc  of  a  great  circle  comprehended  between  two  given  points  on 
the  surface  of  a  sphere  is  less  than  any  arc  of  any  small  circle  compre- 
hended between  the  same  two  points. 

This  follows  from  the  last  theorems. 

THEOREM. 

The  shortest  path  from  one  point  to  another  on  the  surface  of  a  sphere 
is  the  arc  of  a  great  circle. 

To  prove  this,  let  it  be  observed  that  the  sphere  is  perfectly  round 
in  all  directions,  so  that  every  section  of  it  made  by  a  plane  is  a  circle. 
This  being  premised,  suppose  an  irregular  line  upon  its  surface  be- 
tween the  two  given  points ;  this  may  be  considered  either  an  arc  of 
a  small  circle,  or  made  up  of  small  portions  of  such  circles.  In  the 
first  case,  it  has  already  been  proved  that  the  arc  of  a  great  circle  be- 
tween the  points  is  shorter  than  this.  In  the  second  case,  arcs  of  great 
circles  between  the  extremities  of  the  portions  are  less  than  these 
portions,  and,  by  the  repetition  of  the  principle  that  one  side  of  a 
spherical  triangle  is  less  than  the  sum  of  the  other  two,  it  may  be 
shown  that  the  arc  of  a  great  circle  between  the  two  given  points  is 
les3  than  the  polygonal  combination  of  arcs  of  great  circles  between 
the  same  points,  so  that  in  both  cases  the  theorem  is  demonstrated. 

straight  line,  which,  prolonged  both  ways,  becomes  a  tangent ;  the 
tangent,  therefore,  shows  the  direction  of  the  curve  at  the  point  of 
contact.  If,  now,  after  the  arc  AM'B  is  turned  over,  it  be  observed 
that  the  direction  of  this  arc  at  the  point  A  is  perpendicular  to  AO", 
while  the  direction  of  AMB  is  perpendicular  to  AO,  it  is  evident  that 
the  latter  arc  will  run  within  the  former. 

By  joining  the  point  O"  with  any  point  of  the  inverted  arc  AM'B, 
and  tlie  point  in  which  this  line  intersects  the  arc  AMB  with  the 
point  O,  it  may  be  shown  that  the  arc  AM'B  is  every  where  diverging 
in  direction  from  the  arc  AMB,  except  at  M,  M'. 


APPENDIX  IV, 


ISOPERIMETRY  ON  THE  SPHERE. 

1.  Prove  that  of  all  spherical  triangles  formed  with  two  given  sides, 
the  greatest  is  that  in  which  the  angle  formed  by  the  given  sides  is 
equal  to  the  sum  of  the  other  two  angles. 

2.  That  of  all  spherical  triangles  formed  with  one  side,  and  the 
perimeter  given,  the  greatest  is  that  in  which  the  undetermined  sides 
are  equal. 

3.  That  of  all  isoperimetrical  spherical  polygons,  the  greatest  is  an 
equilateral  polygon. 

4.  That  of  all  spherical  polygons  formed  with  given  sides,  and  one 
side  taken  at  pleasure,  the  greatest  is  that  which  can  be  inscribed  in 
a  circle,  of  which  the  chord  of  the  undetermined  side  is  the  diameter. 

5.  The  greatest  of  spherical  polygons  formed  with  given  sides  is 
that  which  can  be  inscribed  in  a  circle  of  the  sphere. 

6.  The  greatest  of  spherical  polygons  having  the  same  perimeter 
and  same  number  of  sides  is  that  in  which  the  sides  and  angles  are 
equal. 

Note.  All  the  above  apply,  also,  to  polyhedral  angles,  of  which  the 
spherical  triangles  are  the  measures. 


APPENDIX  V. 


SYMMETRY    IN    SPACE. 

There  are  two  kinds  of  symmetry  for  polyhedrons,  symmetry  of 
form  and  symmetry  of  position. 

To  give  an  idea  of  these  two  kinds 
of  symmetry,  let  us  consider,  first,  a  tet- 
rahedron SABC,  and  upon  its  edges, 
prolonged  above  the  vertex  S,  take  dis- 
tances SA'  =  SA,  SB'  =  SB,  SC  == 
SC,  and  draw  A'B',  A'C,  B'C ;  the 
parts  of  the  two  tetrahedrons  (edges, 
faces,  diedral  angles)  are  evidently 
equal  each  to  each,  but  disposed  in  an 
inverse  order.  They  are  called  sym- 
metric. 

The  second  tetrahedron  may  be  de- 
tached from  the  first,  and  is  still  symmetric,  whatever  may  be  their 
relative  position. 

Two  polyhedrons  are  said  to  be  symmetric  [and  that  independent 
of  their  position  in  space]  when  they  can  be  decomposed  into  the  same 
number  of  tetrahedrons  symmetric  each  to  each,  and  disposed  in  an  in- 
verse order. 

Whence  it  follows  that,  1°.  A  polyhedron  can  have  but  one  sym- 
metric with  it.  2°.  Two  symmetric  polyhedrons  have  their  edges, 
faces,  diedral  and  polyhedral  angles  equal  each  to  each. 


SYMMETRY   OF    POSITION. 


This  exists  in  three  ways  :  1°.  With  reference  to  a  point,  which  is 
called  a  center  of  symmetry ;  2°.  With  reference  to  a  line,  called  an 
axis  of  symmetry ;  3°.  With  reference  to  a  plane,  called  the  plane  of 
symmetry.      We  shall  treat,  first,  of 


SYMMETRY   RELATIVE   TO   AN   AXIS. 


Definition.  Two  points  are  symmetrical  with  respect  to  a  line 
when  the  line  which  joins  them  is  perpendicular  to  the  first,  and 
divided  by  it  into  two  equal  parts. 


2  GEOMETRY. 

A  polyhedron  is  symmetric,  or  two  polyhedrons  are  symmetric  with 
reference  to  a  line,  when  this  line  passes  through  the  middle  point  of 
all  the  lines  [other  than  the  edges  or  diagonals  of  the  faces]  which 
join  the  vertices  of  the  polyhedron,  two  and  two,  and  is  perpendicular 
to  them. 

Theorem  1.  Two  figures  which  are  symmetric  with  reference  to  a 
line  are  identical. 

This  may  be  proved  by  revolving  the  perpendiculars  about  the  axis ; 
the  vertices  will  all  describe  similar  arcs. 

Corollaries.  In  a  polyhedron  symmetric  with  reference  to  an  axis, 
1°.  Every  line  meeting  the  axis  at  right  angles,  and  terminating  at  the 
surface,  is  equally  divided  by  the  axis.  2°.  Every  plane  through  Ike 
axis  cuts  the  polyhedron  into  tioo  equal  parts.  3°.  Every  plane  per- 
pendicular to  the  axis  determines  a  symmetric  section  with  reference  to 
the  point  of  intersection  of  this  plane  with  the  axis,  and  this  point  is  the 
center  of  symmetry  of  the  section. 

Schol.  1.  The  most  simple  of  polyhedrons  symmetrical  with  refer- 
ence to  an  axis  is  the  right  prism,  the  base  of  which  is  symmetric  with 
reference  to  a  point. 

When  the  base  of  the  right  prism  is  a  rectangle  it  has  for  axes  of 
symmetry  the  three  lines  which  join  the  centers  of  the  opposite  faces. 

If,  moreover,  the  base  is  a  square,  there  exist  two  other  axes  of 
symmetry  which  join  the  middle  of  the  opposite  edges. 

When  the  base  of  the  right  prism  is  a  rhombus,  there  are  three  axes 
of  symmetry,  one  joining  the  centers  of  the  two  bases,  and  two  others 
joining  the  middle  points  of  the  opposite  edges. 

Schol.  2.  The  axis  of  a  regular  pyramid  is  also  an  axis  of  symmetry 
when  the  number  of  lateral  faces  is  even. 

Schol.  3.  Symmetry,  with  reference  to  an  axis,  is,  properly  speak- 
ing, merely  symmetry  of  position,  since,  by  the  preceding  theorem, 
the  figures  are  equal  and  capable  of  superposition.  But  the  same  is 
not  the  case  with  symmetry  with  reference  to  a  point,  or  symmetry 
with  reference  to  a  plane,  which  are  at  the  same  time  symmetry  of 
form  and  position.  For  this  reason  we  have  commenced  with  sym- 
metry referred  to  a  line. 

SYMMETRY   WITH  REFERENCE   TO  A  POINT  OR  PLANE. 

Definitions.  Two  points  are  said  to  be  symmetrical  with  reference 
to  a  point  when  the  latter  divides  into  two  equal  parts  the  line  join- 
ing the  two  former;  and,  with  reference  to  a  plane,  when  this  plane 
is  perpendicular  to  the  line  which  joins  the  two  points  and  bisects  it. 

Theorem  2.   If  three  points  are  in  a  right  line,  their  symmetric 


APPENDIX    V.  3 

points  with  reference  to  a  point  or  plane  arc  in  a  right  line.     The 
student  will  easily  prove  this. 

Corollaries.  1°.  Two  lines  of  determinate  length,  and  symmetric 
with  respect  to  a  point,  are  equal  and  parallel.  2°.  Two  triangles  sym- 
virtric  with  respect  to  a  point  are  equal  and  their  planes  parallel.  3°. 
Two  lines  of  determinate  length,  and  symmetric  with  reference  to  a 
plane,  arc  equal,  make  equal  angles  with  this  plane,  and,  being  prolong- 
ed, meet  it  at  the  same  point,  unless  they  are  parallel. 

Theorem  3.  If  four  points  are  in  the  same  plane,  their  symmetric 
points,  with  reference  to  a  point,  are  also  in  a  same  plane. 

Schol.  When  the  four  points  are  in  different  planes  their  symmetric* 
are  also,  and  then  the  two  systems  of  points  determine  two  tetra- 
hedrons, whose  angles,  diedral  and  trihedral,  are  symmetric,  and, 
consequently,  the  tetrahedrons  themselves  symmetric. 

Theohkm  4.  When  two  polyhedrons  have  their  vertices,  two  and  two, 
si/ m metric  with  reference  to  a  point  or  a  plane  [in  which  case  the  poly- 
hedrons are  said  to  be  symmetric],  1°.  These  polyhedrons  have  their 
faces  equal  each  to  each,  their  diedral  angles  equal  each  to  each,  and 
their  polyhedral  angles  symmetric.  2°.  These  polyhedrons  are  sym- 
metric in  form. 

The  fust  part  of  this  theorem  results  from  the  last  corollaries,  and 
the  second  by  observing  that  the  two  polyhedrons  are  composed  of 
the  same  number  of  tetrahedrons  symmetric,  two  and  two,  and  in- 
versely disposed. 

Thkokkm  5.  When  the  vertices  of  a  polyhedron  are  situated  symmet- 
rically with  reference  to  a  point,  1°.  This  polyhedron  has  necessarily 
an  even  number  of  edges,  equal  and  parallel  two  and  two ;  and  it.  is  the 
same  with  the  faces  ;  2°.  The  plane  angles  and  diedral  angles  arc  also 
equal  each  to  each  ;  the  polyhedral  angles  are  symmetric  in  pairs;  3°. 
Every  line  passing  through  the  center  of  symmetry  and  terminating  at 
the  surface  is  divided  at  this  point  into  two  equal  parts  ;  4°.  Finally, 
every  plane  passing  through  the  center  divides  the  polyhedron  symmetric- 
ally. 

This  follows  from  the  last  corollaries  and  scholium. 

Schol.  1.  The  most  simple  of  polyhedrons  with  reference  to  a  point 
is  the  parallelopipedon.  It  has  for  a  center  of  symmetry  its  center  of 
figure.  As  every  diagonal  plane  passes  through  its  center  of  figure, 
such  a  plane  divides  the  parallelopipedon  symmetrically. 

Schol.  2.  After  the  parallelopipedon,  the  most  simple  are  prisms 
having  for  bases  polygons  symmetric  with  reference  to  a  point.  The 
center  of  symmetry  is  the  middle  of  the  line  which  joins  the  centers 
of  the  two  bases. 


4  GEOMETRY. 

General  Scholium  upon  symmetry  with  reference  to  a  point  and  a 
plane  compared  with  absolute  symmetry. 

It  follows,  from  theorems  four  and  five,  that  two  polyhedrons  sym- 
metric with  reference  to  a  point  or  to  a  plane  are  at  the  same  time 
absolutely  symmetric. 

Reciprocally,  two  polyhedrons  symmetric  to  each  other  (absolutely) 
can  always  be  placed  symmetrically  with  reference  to  a  point  in  space, 
or  with  reference  to  a  plane,  this  point  or  plane  being  a  common  ver- 
tex or  face  of  the  two  polyhedrons.* 

Theorem  6.    Two  symmetric  polyhedrons  are  equivalent. 

It  is  only  necessary,  after  the  first  definition,  to  demonstrate  this 
for  tetrahedrons.  These  may  be  shown  to  have  the  same  base  and 
height  (see  th.  5,  3°,  of  this  App.),  and  are,  consequently,  equal. 

The  two  following  propositions  may  be  easily  established : 

1°.  When  there  exist  in  a  polyhedron  two  planes  of  symmetry  per- 
pendicular to  each  other,  their  common  intersection  is  an  axis  of  sym- 
metry ;  2°.  And  if  there  exist  three,  the  point  common  to  these  three 
planes  is  a  center  of  symmetry. 

OF  DIAMETRAL   PLANES. 

When  a  plane  passes  through  a  polyhedron  in  such  a  manner  that 
a  system  of  parallel  lines  terminating  at  the  surface  are  equally  divided 
by  the  plane,  it  is  called  a  diametral  plane.  N.B. — The  parallels  are 
not  necessarily  perpendicular  to  the  plane. 

Theorem  7.  When  the  vertices  of  a  polyhedron,  or  of  two  polyhe- 
drons, are  situated  in  pairs  upon  parallel  lines,  and  a  certain  plane 
passes  through  the  middle  points  of  these  lines,  1°.  Each  couple  of  ho- 
mologous edges  produced  will  meet  at  a  point  of  the  plane,  unless  they 
are  parallel ;  2°.  Each  couple  of  homologous  planes  determined  by  three 
vertices  of  the  one  polyhedron  and  three  corresponding  of  the  other,  in- 
tersect each  other  in  a  line  of  the  first-mentioned  plane  (unless  they  are 
parallel);  3°.  Every  line  parallel  to  any  of  the  lines  joining  the  ho- 
mologous vertices,  and  terminating  on  either  side  the  plane  at  the  poly- 
hedral surface,  is  equally  divided  by  this  plane,  which  is,  consequently, 
a  diametral  plane. 

N.B. — When  the  lines  joining  the  homologous  vertices  are  equal  and 
parallel,  the  figures  determined  by  the  vertices  are  equal  and  then' 
planes  parallel. 

*  An  object  and  its  reflected  image  present  a  familiar  example  of 
two  figures  symmetric  to  each  other. 

The  human  body  is  a  figure  composed  of  two  parts  symmetric,  with 
reference  to  what  is  called  a  median  plane. 


APPENDIX   V.  O 

Sckol.  1.  The  preceding  theorem  comprehends,  as  a  particular 
case,  figures  symmetrical  with  reference  to  a  plane. 

Sckol.  2.  Every  triangular  prism,  right  or  oblique,  has  four  diame- 
tral planes,  one  of  which  is  the  plaue  half  way  between  the  bases  par- 
allel to  them  ;  and  the  three  others  are  the  planes  passing  through  the 
lateral  edges  and  through  the  diameters  of  the  bases. 

CENTER  OF  MEAN   DISTANCES. 

The  point  which  has  been  named  center  of  mean  distances  in  a  pol- 
ygon in  a  previous  appendix  (II.,  def.  4),  has  a  property  with  refer- 
ence to  a  plane  which  we  have  shown  it  to  have  with  reference  to  a 
line. 

Theorem  8.  The  perpendicular  let  fall  from  the  center  of  mean  dis- 
tances upon  a  plane  drawn  at  pleasure  in  space,  is  equal  to  the  quotient 
of  the  algebraic  sum  of  the  perpendiculars  let  fall  from  the  different  ver- 
tices upon  this  plane  divided  by  the  number  of  vertices.  This  sum  is 
zero  when  the  plane  passes  through  the  center  of  mean  distances,  and 
vice  versa. 

The  demonstration  will  be  similar  to  that  in  the  corresponding  one 
in  a  previous  appendix  (II.,  def.  4,  et  seq.). 

It  is  to  be  observed,  that  the  vertices  of  which  the  point  in  question 
is  the  center  of  mean  distances  need  not  be  in  the  same  plane  as  they 
were  supposed  to  be  in  the  previous  appendix. 

Schol.  To  determine  the  center  of  mean  distauces  for  any  number 
of  points  not  in  the  same  plane,  draw  three  planes  at  pleasure  which 
cut  each  other  (suppose,  for  the  sake  of  simplicity,  at  right  angles). 
Let  fall,  from  the  different  vertices  upon  each  of  these  planes,  per- 
pendiculars ;  find  afterward  for  each  plane  the  algebraic  sum  of  its 
perpendiculars,  and  divide  this  sum  by  the  number  of  vertices.  Fi- 
nally, at  distances  equal  to  the  three  quotients,  draw  planes  parallel 
to  the  first  three,  and  their  common  intersection  will  be  the  point 
sought. 

When  four  points  are  not  in  the  same  plane,  these  points,  combined 
three  and  three,  determine  a  tetrahedron.     This  being  observed : 

Theorem  9.  In  every  tetrahedron  the  lines  which  join  the  middle 
points  of  the  edges  not  adjacent,  all  meet  in  a  point  which  is  the  center 
of  mean  distances  of  the  four  vertices. 

Schol.  This  point  is  also  found  in  three  planes  parallel  to  the  faces, 
and  at  a  distance  equal  to  one  quarter  the  distance  of  the  opposite 
vertex  from  each  face. 

Theorem  10.  The  four  lines  joining  the  vertices  with  the  centers  of 
mean  disla?ices  of  the  opposite  faces,  meet  in  a  point  which  is  the  cen- 


6  GEOMETRY. 

ter  of  mean  distances  of  the  vertices.  This  point  is  one  quarter  the  dis- 
tance from  the  center  of  mean  distances  in  each  face  to  the  opposite 
vertex. 

OF    CENTERS    OF    SIMILITUDE. 

Theorem  11.  If  all  the  vertices  of  a  polyhedron  be  joined  with  a 
point  in  space  by  lines,  and  upon  these  lines,  or  three  prolongations,  por- 
tions be  taken  proportional  to  the  lines  themselves,  the  vertices  of  a  new 
polyhedron  will  be  thus  obtained,  which  is  directly  or  inversely  similar  to 
the  first. 

This  point  is  called  a  center  of  similitude,  external  in  the  first 
case,  internal  in  the  second. 

The  proof  of  the  above  is  in  all  respects  similar  to  that  in  a  previ- 
ous appendix  (App.  II.). 

CENTERS   OF   SIMILITUDE    OF   SPHERES. 

If  lines  be  drawn  tangent  to  two  circles  meeting  each  other,  one 
pair  internally  and  the  other  pair  externally ;  and  if  these  circles  and 
tangents  be  set  in  revolution  about  the  line  joining  the  centers,  the 
circles  will  generate  spheres,  and  the  tangents,  cones  enveloping  the 
spheres,  and  the  points  of  contact  will  generate  circles  which  will  be 
the  curves  of  contact  of  the  cones  and  spheres ;  the  planes  of  these 
circles  of  contact  will  be  perpendicular  to  the  axis.  The  vertices  of 
these  cones,  at  the  points  in  which  the  tangents  intersect,  are  called 
centers  of  similitude  of  the  two  spheres,  the  one  internal,  the  other 
external. 

Prove  that  every  plane  tangent  to  one  of  these  conic  surfaces  is  tan- 
gent to  the  two  spheres. 

And,  conversely,  that  every  plane  tangent  to  the  two  spheres  is  tan- 
gent to  one  of  the  conic  surfaces. 

Two  spheres  in  space  would  have  an  infinite  number  of  common 
tangent  planes.  One  of  these  would  be  determined  by  another  con- 
dition, as,  that  it  should  pass  through  a  given  point,  or  be  parallel  to  a 
given  line,  or  tangent  to  a  third  sphere,  &c. ;  and  there  would  be  two 
planes  which  would  fulfill  the  required  condition  in  the  first  two  cases ; 
in  the  last  there  might  be  four  systems  of  two  planes  tangent  to  the 
three  spheres,  to  wit :  two  planes  comprehending  the  three  spheres 
between  them,  and  six  placed  two  and  two  between  one  of  the 
spheres  and  the  two  others. 

This  second  case  gives  rise  to  a  remarkable  theorem  analogous  to 
one  in  a  previous  appendix  (App.  II.),  for  three  circumferences  of  a 
circle. 


APPENDIX  V.  7 

Theorem  12.    The  six  centers  of  similitude  of  three  spheres  ea 
to  one  another  are  situated  three  and  three  upon  a  same  line,  to 
wit,  the  three  external  centers  of  similitude,  then  one  of  the  external 
and  two  internal,  giving  in  all  four  lines. 

For,  first,  let  us  consider  the  two  tangent  planes  which  embrace  the 
three  spheres  between  them.  These  planes  being  tangent  to  the  three 
cones  which  envelop  the  spheres,  must  both  pass  through  the  vertices 
of  these  cones,  and,  consequently,  their  intersection  must.  The  other 
tangent  planes  will,  in  a  similar  manner,  serve  to  demonstrate  the 
other  part  of  the  theorem. 

This  theorem  serves  to  prove  the  correctness  of  the  theorem  for  the 
case  of  three  circumferences  (App.  II.),  because  the  centers  of  simil- 
itude of  these  circles  are  the  same  as  the  centers  of  similitude  of  three 
spheres,  of  which  these  circles  are  great  circles. 

It  is  thus  that  sometimes  propositions  in  Plane  Geometry  may  be 
demonstrated  in  a  more  simple  manner  by  the  aid  of  truths  relating 
to  geometry  in  space. 


REGULAR  POLYHEDRONS. 

A  regular  polyhedron  is  one  in  ichich  the  faces  are  equal  regular  pol- 
ygons, and  the  diedral  angles  equal.  From  this  definition  it  will  follow 
that  the  polyhedral  angles  will  also  be  equal. 

THEOREM. 

There  can  be  but  five  regular  polyhedrons. 

This  follows  from  Prop.  2,  of  Polyhedral  Angles,  that  a  polyhedral 
angle  can  not  be  formed  unless  the  sum  of  the  plane  angles  which 
form  it  is  less  than  four  right  angles. 

If  we  take  equilateral  triangles,  each  angle  of  which  is  two  thirds 

of  a  right  angle,  to  form  a  polyhedral  angle,  we  may  combine  these 

2       12 
in  threes,  fours,  and  fives,  but  not  more,  because  6  X  -  =  -%■  =  4  right 

o         o 

angles. 

If  we  take  squares,  each  angle  of  which  is  one  right  angle,  to  form 

a  polyhedral  angle,  we  can  combine  them  in  threes  alone,  for  4  X  1  = 

4  right  angles. 

5^2 4 

If  regular  pentagons,  each  angle  of  which  is =  1 1,  they 

can  be  combined  but  in  threes. 

If  hexagons,  each  angle  of  which  is  \\,  they  can  not  be  combined 


8  GEOMETRY 

even  in  threes  to  form  a  polyhedral  angle,  and  three  is  the  least  num- 
ber of  planes  that  can  be  employed  for  this  purpose. 

It  is  evident  that  still  less  can  regular  polygons  of  a  greater  number 
of  sides  be  employed. 

There  can,  therefore,  be  formed  but  three  regular  polyhedrons  of 
triangular  faces,  but  one  of  square  faces,  and  but  one  of  pentagonal 
faces,  in  all  five,  which  is  the  greatest  number  that  can  possibly  exist. 

Schol.  It  remains  to  be  shown  that  five  regular  polyhedrons  can  be 
formed. 

CONSTRUCTION  OF  REGULAR  POLYHEDRONS. 

1°.   TO   CONSTRUCT   A  REGULAR   TETRAHEDRON. 

Take  an  eqtiilateral  triangle ;  erect  at  the  center  of  its  inscribed  cir- 
cle a  perpendicular  to  its  plane ;  with  one  of  its  vertices  as  a  center, 
and  a  radius  equal  in  length  to  one  of  its  edges,  cut  this  perpendicu- 
lar in  a  point ;  join  this  point  with  the  vertices  of  the  triangle,  and  the 
regular  tetrahedron  will  be  formed. 

2°.    TO   CONSTRUCT  A  REGULAR   HEXAHEDRON   OR   CUBE. 

We  leave  this  to  the  student,  being  too  easy  to  require  explanation. 

3°.   A   REGULAR  OCTAHEDRON. 

Upon  a  line  equal  to  one  of  the  sides  of  the  equilateral  triangle, 
which  is  to  be  a  face,  construct  a  square ;  erect  at  the  center  of  this 
square  a  perpendicular  to  its  plane,  and  take  upon  this  perpendicular, 
on  each  side  of  the  plane,  a  distance  equal  to  one  half  the  diagonal  of 
the  square ;  joining  the  points  thus  determined  with  the  vertices  of 
the  square,  the  polyhedron  required  is  formed. 

N.B. — The  center  of  the  square  is  a  center  of  symmetry.  It  is  also 
the  center  of  figure. 

4°.   A   REGULAR  ICOSAHEDRON. 

Construct  first  a  pentagon  upon  the  side  of  the  given  equilateral 
triangle ;  at  the  center  of  this  figure  erect  a  perpendicular  to  its 
plane  ;  with  a  radius  equal  to  the  side  of  the  triangle,  cut  this  per- 
pendicular in  a  point ;  this  point  being  joined  with  the  vertices  of  the 
pentagon,  will  furnish  five  equilateral  triangles  formed  about  it ;  form 
now  a  second  pentahedral  angle,  with  one  of  the  angles  of  the  pen- 
tagon as  a  vertex,  and  two  of  its  faces  will  be  the  same  as  those  of  the 
first  pentahedral  angle  formed ;  with  a  third  vertex  of  the  same  trian- 
gle, to  which  the  other  two  already  employed  belonged,  form  a  third 


APPENDIX   V.  9 

pentahedral  angle;  for  this  purpose  two  new  faces  will  be  required. 
There  will  thus  be  united  ten  triangles,  forming  a  sort  of  polyhedral 
cap,  such  that  the  angles  at  the  border  are  formed  by  alternately  two 
and  three  triangles.  This  polygonal  line,  winch  terminates  the  surface, 
has  its  sides  equal,  but  its  vertices  not  in  the  same  plane.  If  now  a 
second  polyhedral  cap  be  constructed  equal  to  the  first,  its  diedral 
angles  will  have  the  same  value  as  those  in  the  other.  Then,  with- 
out breaking  the  continuity,  we  can  unite  the  double  angles  of  the  bor- 
der of  the  first  cap  with  the  triple  angles  of  the  border  of  the  second, 
and  vice  versa;  whence  will  result  a  figure  of  twenty  equal  faces 
equally  inclined. 

5°.    A  REGULAR  DODECAHEDRON. 

Suppose  that  with  three  regular  pentagons  a  trihedral  be  formed, 
which  is  possible  (see  last  th.).  The  three  diedral  angles  of  this  tri- 
hedral angle  are  equal.  Now  with  new  pentagons,  equal  to  the  pre- 
ceding, can  be  formed  in  the  same  manner,  successively,  at  the  vertices 
of  one  of  these  pentagons,  other  trihedral  angles,  all  of  the  same  mag- 
nitude. There  will  result  six  regular  pentagons,  composing  a  polyg- 
onal cap,  such  that  the  angles  of  the  border  are  formed  alternately 
of  one  arid  of  two  plane  angles. 

[The  same  remark  as  above  applies  to  this  border.] 

If  a  second  cap  be  imagined,  equal  to  the  first,  they  can  be  united, 
border  to  border,  so  that  the  single  angles  of  the  one  accord  to  the 
double  angles  of  the  other ;  and  thus  will  be  formed  a  figure  of 
twelve  faces,  equal,  and  equally  inclined  to  one  another. 

Schol.  1.  To  construct  a  regular  polyhedron  mechanically,  taking 
one  of  the  faces  as  a  base  of  construction,  upon  a  sheet  of  pasteboard 
make  the  development  of  all  the  faces,  then  fold  these  different  faces 
upon  their  edges  in  a  suitable  manner. 

Schol.  2.  All  the  regular  polyhedrons  except  the  tetrahedron  have 
a  center  of  symmetry  which  is  identical  with  the  center  of  figure. 

All  have,  also,  planes  of  symmetry.  These  are,  in  general,  planes 
perpendicular  upon  the  middle  of  the  edges,  or  upon  the  middle 
points  of  lines  joining  opposite  vertices,  taken  two  and  two,  or  else 
planes  passing  through  the  opposite  edges,  two  and  two. 

Schol.  3.  The  regular  tetrahedron  has  4  vertices,  4  faces,  and  6  edges. 
The  cube  8       "        6     "  12     " 

The  octahedron  6       "         8     "  12     " 

The  dodecahedron  20       "       12     "  30     " 

The  icosahedron  12       "      20     "  30     " 

General  Scholium  upon  Polyhedrons.   These  expressions,  which  can 


10  GEOMETRY. 

be  easily  verified  upon  the  figures,  are  contained  in  the  enunciation 
of  a  theorem  by  the  celebrated  Euler,  and  which  is  translated  by  the 
formula 

V-f-F  =  E-f2; 
V  designating  the  number  of  vertices,  F  the  number  of  faces,  and  E 
the  number  of  edges.     This  formula  has  been  previously  given. 

There  are  a  great  many  theorems  more  or  less  important  upon 
polyhedrons  as  well  as  upon  polygons,  similar  to  those  in  a  previous 
appendix,  for  which,  see  a  memoir  of  M.  Poinsot,  in  the  Journal  de 
l'Ecole  Polytechnique,  10e  cahier,  t.  iv.,  p.  6,  et  seq.  Also,  a  memoir 
of  M.  Cauchy,  in  the  same  journal,  16e  cahier,  t.  ix.,  p.  77.  Also, 
Annales  de  Mathematiques  of  M.  Gergonne,  particularly  tome  xv., 
page  157. 


MENSURATION  OF  PLANES. 


D  4  O 


The  area  of  any  plane  figure  is  the  measure  of  the 
space  contained  within  its  extremes  or  bounds,  with- 
out any  regard  to  thickness. 

This  area,  or  the  content  of  the  plane  figure,  is  es- 
timated by  the  number  of  little  squares  that  may  be 
contained  in  it ;  the  side  of  each  of  those  little  meas- 
uring squares  being  an  inch,  a  foot,  a  yard,  or  any 
other  fixed  quantity.  And  hence  the  area  or  content 
is  said  to  be  so  many  square  inches,  or  square  feet,  or 
square  yards,  &c.  In  other  words,  the  area  of  a  sur- 
face is  the  numerical  ratio  of  this  surface  to  its  unit. 

Thus,  if  the  figure  to  be  measured 
be  the  rectangle  ABCD,  and  the  little 
square  E,  whose  side  is  one  inch,  be 
the  measuring  unit  proposed  ;  then,  as 
often  as  the  said  little  square  is  con- 
tained in  the  rectangle,  so  many  square 
inches  the  rectangle  is  said  to  contain, 
which  in  the  present  case  is  12. 


PROBLEM    I. 

To  find  the  area  of  any  parallelogram,  whether  it  be 
a  square,  a  rectangle,  a  rhombus,  or  a  rhomboid. 

Multiply  the  length  by  the  perpendicular  breadth 
or  height,  and  the  product  will  be  the  area.* 

*  The  truth  of  this  rule  is  proved  in  the  Geometry,  Theor.  60, 
Schol. 

The  same  is  otherwise  proved  thus :  Let  the  foregoing  rectangle  be 
the  figure  proposed ;  and  let  the  length  and  breadth  be  divided  into 
equal  parts,  each  equal  to  the  linear  measuring  unit,  being  here  four 
for  the  length  and  three  for  the  breadth;  and  let  the  opposite  points 
of  division  he  connected  by  right  lines.  Then  it  is  evident  that  these 
lines  divide  the  rectangle  into  a  number  of  little  squares,  each  equal 
to  the  square  measuring  un't  E ;  and  further,  that  the  number  of  these 
I 


MENSURATION. 


EXAMPLES. 


Ex.  1.  To  find  the  area  of  a  parallelogram  whose 
length  is  1225,  and  height  8*5. 

12-25  length. 
8-5  breadth. 

6125 

9800 


104*125  area. 

Ex.  2.  To  find  the  area  of  a  square  whose  side 
is  35*25  chains.       Ans.  124  acres,  1  rood,  1  perch. 

Ex.  3.  To  find  the  area  of  a  rectangular  board 
whose  length  is  12^  feet,  and  breadth  9  inches. 

Ans.  9 1  feet. 

Ex.  4.  To  find  the  content  of  a  piece  of  land  in 
form  of  a  rhombus,  its  length  being  6*20  chains,  and 
perpendicular  height  5*45. 

Ans.  3  acres,  1  rood,  20  perches. 

Ex.  5.  To  find  the  number  of  square  yards  of 
painting  in  a  rhomboid  whose  length  is  37  feet,  and 
breadth  5  feet  3  inches.      Ans.  21T72  square  yards. 

PROBLEM   II. 

To  find  the  area  of  a  triangle. 

Rule  I.  Multiply  the  base  by  the  perpendicular 
height,  and  half  the  product  will  be  the  area.*  Or, 
multiply  the  one  of  these  dimensions  by  half  the 
other. 

little  squares,  or  the  area  of  the  figure,  is  equal  to  the  number  of  linear 
measuring  units  hi  the  length,  which  is  the  same  as  the  number  of 
square  units  in  a  horizontal  row,  repeated  as  often  as  there  are  linear 
measuring  units  in  the  breadth  or  height,  which  is  the  same  as  the 
number  of  horizontal  rows,  that  is  here  4  X  3  or  12. 

And  it  is  proved  (Geometry,  theor.  22)  that  a  rectangle  is  equal  to 
any  oblique  parallelogram  of  equal  length  and  perpendicular  breadth. 
Therefore  the  rule  is  general  for  all  parallelograms  whatever. 

*  The  truth  of  this  rule  is  evident,  because  any  triangle  is  the  half 
of  a  parallelogram  of  equal  base  and  altitude,  by  Geometry,  Theor. 
23. 


MENSURATION    OP    PLANES. 


EXAMPLES. 


Ex.  1.  To  find  the  area  of  a  triangle  whose  base 
is  625,  and  perpendicular  height  520  links  ?# 

Here         625  X  260=  162500  square  links, 
or  equal  1  acre,  2  roods,  20  perches,  the  answer. 

Ex.  2.  How  many  square  yards  contains  the  tri- 
angle, whose  base  is  40,  and  perpendicular  30  feet  ? 

Ans.  66§  square  yards. 

Ex.  3.  To  find  the  number  of  square  yards  in  a 
triangle  whose  base  is  49  feet,  and  height  25|  feet. 

Ans.  68ff,  or  68-7361. 

Ex.  4.  To  find  the  area  of  a  triangle  whose  base 
is  18  feet  4  inches,  and  height  11  feet  10  inches. 

Ans.  108  feet,  5|  inches. 

Rule  II.  When  two  sides  and  their  contained  an- 
gle are  given  :  Multiply  the  two  given  sides  together, 
and  take  half  their  product :  Then  say,  as  radius  is  to 
the  sine  of  the  given  angle,  so  is  that  half  product  to 
the  area  of  the  triangle. 

Or,  multiply  that  half  product  by  the  natural  sine 
of  the  said  angle,  f 

Ex.  1.  What  is  the  area  of  a  triangle  whose  two 
sides  are  30  and  40,  and  their  contained  angle  28°  57' 
18"? 

Here  \  x  40  X  30  =  600, 

Therefore,     -4841226  nat.  sin.  28°  57'  18" 
600 

290*47356,  the  answer. 
Ex.  2.  How  many  square  yards  contains  the  tri- 

*  100  links  make  a  chain,  10,000  square  links  a  square  chain,  and 
10  square  chains  an  acre. 

t  The  following  demonstration  requires 
an  acquaintance  with  Trigonometry.  For, 
let  AB,  AC  be  the  two  given  sides,  includ- 
ing the  given  angle  A.  Now  £AB  X  CP 
is  the  area,  by  the  first  rule,  CP  being  per- 
pendicular. But,  by  Trigonometry,  CP 
t=  sine  angle  A  X  AC,  taking  radius  =  1.  Therefore,  the  area  £AB 
X  CP  is  =£AB  x  AC  X  sin.  angle  A,  to  radius  1 ;  or, 

as  radius :  sin.  angle  A : :  JAB  x  AC  :  the  area. 


4  MENSURATION. 

angle,  of  which  one  angle  is  45°,  and  its  containing 
sides  25  and  21 1  feet?  Ans.  20*86947. 

Rule  III.  When  the  three  sides  are  given  :  Add  all 
the  three  sides  together,  and  take  half  that  sum. 
Next,  subtract  each  side  severally  from  the  said  half 
sum,  obtaining  three  remainders.  Lastly,  multiply 
the  said  half  sum  and  those  three  remainders  all  to- 
gether, and  extract  the  square  root  of  the  last  prod- 
uct for  the  area  of  the  triangle.* 

Ex.  1.  To  find  the  area  of  the  triangle  whose 
three  sides  are  20,  30,  40. 

20  45  45  45 

30  20  30  40 

40  25,  first  rem.   15,  second  rem.     5,  third  rem. 

2)90  A 

45,  half  sum. 

Then  45  X  25  X  15  X  5  =  84375. 

The  root  of  which  is  290*4737,  the  area. 

Ex.  2.  How  many  square  yards  of  plastering*  are 
in  a  triangle  whose  sides  are  30,  40,  50  ? 

Ans.  66|. 

Ex.  3.  How  many  acres,  &c,  contains  the  triangle 
whose  sides  are  2569,  4900,  5025  links  ? 

Ans.  61  acres,  1  rood,  39  perches. 

PROBLEM.  III. 

To  find  the  area  of  a  trapezoid. 
Add  together  the  two  parallel  sides  ;  then  multiply 

*  For,  let  a,  b,  c  denote  the  sides  opposite  respectively  to  A,  B,  C, 
the  angles  of  the  triangle  ABC  (see  last  figure)  ;  then,  by  theor. 
29,  Geom.,  we  have   BC2  — AB2-f  AC2  — 2AB.AP,  or   a2  =  i2-j- 

c2  — 2c.  AP  .-.  AP=  — £| ;  hence  we  have 

npa—  p     (&2+c2— a2)2_4&2c2— (b*+c*—a*y    (<&<>+■&+&— a*)' (2bc—b*—c*+a,z) 

4c2  4c2  ic2 

.\4c2.CP2=  j  (&4-c)2_a2  j .  J  a2— (c— 6)2  j  =(a+6+c)  (.^^^^  (a-b+c)  (a+b—c) 

•  1XR  PP-!r  PP—  A  <*+H-c  —a+b+c  a—b+c  a+b—c }    

..-AJ3.CP--c.CF_v/l_ _ m  ______  }_V,(»-flX_4)(_c) 

where  t  =„(a+b-\-c)  —  half  the  sum  of  the  three  sides. 


MENSUBATION   OF   (LANES.  5 

their  sum  by  the  perpendicular  breadth  or  distance 
between  them;  and  half  the  product  will  be  the  area, 
by  Geometry,  theorem  25. 

.  l.  In  a  trapezoid  the  parallel  sides  are  750 
and   1225,  and  the   perpendicular   distance  between 
them  1510  links:  to  find  the  area. 
I 

750 
1975  X  770  =  152075  sq.  links  =  15  acres,  33  perches. 

Ex.  2.  How  many  square  feet  are  contained  in  the 
plank  whose  length  is  12  feet  6  inches,  the  breadth  at 
lhe  greater  end  15  inches,  and  at  the  less  end  11 
inches?  Ans.   13|f  feet. 

Ex,  3.  In  measuring  along  one  side  AB  ofa  quad- 
,!ar  field,  that  side  and  the  two  perpendiculars 
let  fall  on  it  from  the  two  opposite  corners,  measured 
as  below  :  required  the  content.  D 

AP  =    110  links. 

A<i=    745     "  C 

AB=1110     "  A  \ 

CP  =    352     "  /!     •  !       \ 

DQ=    595     "  AP  Q        B 

Ans.  4  acres,  1  rood,  5*792  perches. 

rROBLH.M   IV. 

To  find  the  area  of  any  trapezium. 

Divide  the  trapezium  into  two  triangles  by  a  diag- 
onal :  then  find  the  areas  of  these  triangles,  and  add 
them  together. 

Note.  If  two  perpendiculars  be  let  fall  on  the  diag- 
onal, from  the  other  two  opposite  angles,  the  sum  of 
perpendiculars  being  multiplied  by  the  diagonal, 
half  the  product  will  be  the  area  of  the  trapezium. 

Ex.  1.  To  find  the  area  of  the  trapezium  whose 
diagonal  is  42,  and  the  two  perpendiculars  on  it  10 
and  18. 

!  Fere  16  +  18  =    34  ;  its  half  is  17. 

Then  42  X  17  =  714,  the  area. 

Ex.  2.  How  many  square  yards  of  paving  are  in 


6  MENSURATION. 

the  trapezium  whose  diagonal  is  65  feet,  and  the  two 
perpendiculars  let  foil  on  it  28  and  331  feet  ? 

Ans.  222T,2  yards. 
Ex.  3.  In  the  quadrangular  field  ABCD,  on  account 
of  obstructions,  there  could  only  be  taken  the  follow- 
ing measures,  viz. :  the  two  sides  BC  265,  and  AD  220 
yards,  the  diagonal  AC  378,  and  the  two  distances  of 
the  perpendiculars  from  the  ends  of  the  diagonal, 
namely,  AE  100,  and  CF  70  yards.  Required  the 
area  in  acres  when  4840  square  yards  make  an  acre. 
Ans.  17  acres,  2  roods,  21  perches. 

problem  v. 
To  find  the  area  of  an  irregular  polygon. 
Draw  diagonals  dividing  the  proposed  polygon  into 
trapeziums  and  triangles.     Then  find  the  areas  of  all 
these  separately,  and  add  them  together  for  the  con- 
tent of  the  whole  polygon. 

Ex.  To  find  the  content  of  the   irregular   figure 
ABCDEFGA,  in  which  are  given  the  following  diag- 
onals and  perpendiculars,  B 
namely :                                                         ^/X 
AC  55                                            ^<       \  \ 
FD52                                       >^  •      \ 

Gm  13  X  ^""$\       \ 

B?i  18  \^'""'^  \    / 

Go  12  /i       *--^        \/ 

D?23  ^^^^A^^ 

Ans.   1878-5.  E 

PROBLEM   VI. 

To  find  the  area  of  a  regular  polygon. 
Rule  I.  Multiply  the  perimeter  of  the  polygon,  or 
sum  of  its  sides,  by  the  perpendicular  drawn 'from  its 
center  on  one  of  its  sides,  and  take  half  the  product 
for  the  area.* 

*  The  demonstration  of  this  is  given  in  th.  73. 


MENSURATION    OF    PLANES.  7 

Ex.  1.  To  find  the  area  of  the  regular  pentagon, 
each  side  being  25  feet,  and  the  perpendicular  from 
the  center  on  each  side  17*2047737. 

Here  25  x    5   =  125,  is  the  perimeter. 

And  17*2047737  x  125  =  2150-5967125. 

Its  half,  1075*298356,  is  the  area  sought. 

Rule  II.  Square  the  side  of  the  polygon  ;  then  mul- 
tiply that  square  by  the  area  or  multiplier  set  against 
its  name  in  the  following  table,  and  the  product  will 
be  the  area.* 


No.  of 

Names. 

Areas  or 

Sides. 
3 

Multipliers. 

Trigon,  or  triangle 

0-4330127 

4 

Tetragon,  or  square 

1-0000000 

5 

Pentagon 

1-7204774 

6 

Hexagon 

2-5980762 

7 

Heptagon 

3-6339124 

8 

Octagon 

4-8284271 

9 

Nonagon 

6-1818242 

10 

Decagon 

7-6942088 

11 

Undecagon 

9-3656399 

1*2 

Dodecagon 

11-1961524 

Ex.    Taking  here  the   same   example  as  before, 
namely,  a  pentagon,  whose  side  is  25  feet. 

Then,  25a  being  -  625, 

And  the  tabular  area  1*7204774; 
Therefore,  1-7204774  x  625=  1075*298375,  as  before. 


*  This  rule  is  founded  on  the  property  that  regular  polygons  of 
the  same  number  of  sides,  being  similar  figures,  are  as  the  squares  of 
their  sides.  Now  the  multipliers  in  the  table  are  the  areas  of  the  re- 
spective polygons  to  the  side  1.     Whence  the  rule  is  manifest. 

Note.  The  areas  in  the  table,  to  each  side  1,  may 
1><-  oqmpated  in  the  following  manner,  with  the  aid 
of  plane  trigonometry :  From  the  center  C  of  the 
polygon  draw  lines  to  every  angle,  dividing  the 
whole  figure  into  as  many  equal  triangles  as  the  pol- 
ygon has  sides  ;  and  let  ABC  be  one  of  those  trian- 
gles, the  perpendicular  of  which  is  CD.  Divide 
360  degrees  by  the  number  of  sides  in  the  polygon,  A  D  B 
the  quotient  gives  the  angle  at  the  center  ACB.  The  half  of  this  gives 
the  angle  ACD;  and  this  taken  from  90°,  leaves  the  angle  CAD. 
Then,  as  radius  is  to  AD,  so  is  tangent  angle  CAD  to  the  perpendicular 
CD.  This,  multiplied  by  AD,  gives  the  area  of  the  triangle  ABC; 
which,  being  multiplied  by  the  number  of  the  triangles,  or  of  the 
sides  of  the  polygon,  gives  its  whole  area,  as  in  the  table. 


8 


MENSURATION. 


Ex.  2.  To  find  the  area  of  the  trigon,  or  equilateral 
triangle,  whose  side  is  20.  Ans.  173-20508. 

Ex.  3.  To  find  the  area  of  a  hexagon  whose  side 
is  20.  Ans.   103923048. 

Ex.  4.  To  find  the  area  of  an  octagon  whose  side 
is  20.  Ans.   1931-37084. 

Ex.  5.  To  find  the  area  of  a  decagon  whose  side 
is  20.  Ans.  3077-68352. 

PROBLEM    VII. 

To  find  the  diameter  and  circumference  of  any  circle, 
the  one  from  the  other. 

This  may  be  done  nearly  by  either  of  the  two  fol- 
lowing proportions,  viz.  : 

As  7  is  to  22,  so  is  the  diameter  to  the  circumfer- 
ence. 

Or,  as  1  is  to  3-1416,  so  is  the  diameter  to  the  cir- 
cumference.* 

*  For,  let  ABCD  be  any  circle  whose  center  is  E, 
and  let  AB,  BC  be  any  two  equal  arcs.  Draw  the 
several  chords  as  in  the  figure,  and  join  BE  ;  also, 
draw  the  diameter  DA,  which  produce  to  F,*  till  BF 
be  equal  to  the  chord  BD. 

Then  the  two  isosceles  triangles  DEB,  DBF  are  equi- 
angular, because  they  have  the  angle  at  D  common ; 
consequently,  DE  :  DB  : :  DB  :  DF.  But  the  two  tri- 
angles AFB,  DCB  are  identical,  or  equal  in  all  respects, 
because  they  have  the  angle  F  =  the  angle  BDC,  be- 
ing each  equal  the  angle  ADB  (see  th.  39,  cor.  1 ) ;  also, 
the  exterior  angle  FAB  of  the  quadrangle  ABCD  is  equal  the  opposite 
interior  angle  at  C  (exercise  32,  p.  48) ;  and  the  two  triangles  have, 
also,  the  side  BF  =  the  side  BD ;  therefore,  the  side  AF  is  also  equal 
the  side  DC.  Hence  the  proportion  above,  viz.,  DE  :  DB : :  DB :  DF 
=  DA  -f-  AF,  becomes  DE  :  DB  : :  DB  :  2DE  +  DC.  Then,  by  taking 
the  rectangles  of  the  extremes  and  means,  it  is  DB2  =  2DE2  -4-  DE  . 
DC 

Now  if  the  radius  DE  be  taken  ==  1,  this  expression  becomes  DB3 

=  2  -f-  DC  ;  and  hence  DB  =  V24-DC.  That  is,  if  the  measure  of 
the  supplemental  chord  of  any  arc  be  increased  by  the  number  2, 
the  square  root  of  the  sum  will  be  the  supplemental  chord  of  half  that 
arc. 

Now,  to  apply  this  to  the  calculation  of  the  circumference  of  the 
circle,  let  the  arc  AC  be  taken  equal  to  one  sixth  of  the  circumference, 

*  The  point  F  may  be  found  by  describing  an  arc  with  B  as  center,  and  radius 
=  BD. 

f  The  supplemental  chord  is  the  chord  of  the  supplement. 


MENSURATION   OF    PLANES. 


0 


for  the 

>  supplemental 

chord  of 


1 

T2 

& 

1U2 

1 

TO* 
TSJVJ 


of  the 
periphery. 


Ex.  1.  To  find  the  circumference  of  the  circle 
whose  diameter  is  20. 

and  be  successively  bisected  by  the  above  theorem :  thus,  the  chord 
AC  of  one  sixth  of  the  circumference  is  the  side  of  the  inscribed  reg- 
ular hexagon,  and  is,  therefore,  equal  the  radius  AE  or  1 ;  hence,  in 
the  right-angled  triangle  ACD,  we  shall  have  DC  =  n/AD1  —  AC*  = 
-v/2'2  —  l3  =  %/  3  =  1-7320508076,  the  supplemental  chord  of  one  sixth 
of  the  periphery. 

Then,  by  the  foregoing  theorem,  by  always  bisecting  the  arcs,  and 
adding  2  to  the  last  square  root,  there  will  be  fouud  the  supplemental 
chords  of  the  12th,  the  24th,  the  48th,  the  96th,  &c.,  parts  of  the 
periphery;  thus, 

^3-7320508076  =  1-9318516525 " 
^3-9318516525  =  1-9828897227 
v/3'9828897227  =  19957 178465 
v/3-9957 178465  =  1-9989291743 
v/3-9989291743  =  1-9997322757 
^3-9997322757  =  1-9999330678 
-v/3-9999330678  =  1-9999832669 
^3-9999832669  = 

Since,  then,  it  is  found  that  3-9999832669  is  the  square  of  the  sup- 
plemental chord  of  the  1536th  part  of  the  periphery,  let  this  number 
be  taken  from  4,  the  square  of  the  diameter,  and  the  remainder 
0-0000167331  will  be  the  square  of  the  chord  of  the  said  1536th  part 
of  the  periphery,  and,  consequently,  the  root  ^000001 67331  = 
0-0040906112  is  the  length  of  that  chord;  this  number,  then,  being 
multiplied  by  1536,  gives  6-2831788  for  the  perimeter  of  a  regular 
polygon  of  1536  sides  inscribed  in  the  circle;  which,  as  the  sides  of 
the  polygon  nearly  coincide  with  the  circumference  of  the  circle, 
must  also  express  the  length  of  the  circumference  itself,  very  nearly. 

But  now,  to  show  how  near  this  determination  is  to  the  a  R  T 
truth,  let  AQP  =  0-0040906112  represent  one  side  of  such 
a  regular  polygon  of  1536  sides,  and  SRT  a  side  of  another 
similar  polygon  described  about  the  circle ;  and  from  the 
center  E  let  the  perpendicular  EQR  be  drawn,  bisecting 
AP  and  ST  in  Q  and  R.  Then,  since  AQ  is  =  iAP  = 
0-0020453056,  and  EA  =  1,  therefore  EQ2  =  EA2  —  AQ* 
=  •9999958167,  aud,  consequently,  its  root  gives  EQ  = 
•9999979084  ;  then,  because  of  the  parallels  AP,  ST,  we 
have  the  proportion  EQ :  ER : :  AP  :  ST : :  the  whole  in- 
scribed perimeter  :  the  circumscribed  one  ;  that  is,  as 
•9999979084  :  1 : :  6-2831788:  6-2831920,  the  perimeter  of 
the  circumscribed  polygon.  But  the  circumference  of  the  circle  be- 
ing greater  than  the  penmeter  of  the  inner  polygon,  and  less  than  that 
of  the  outer,  it  must,  consequently,  be  greater  than  6-2831788, 

but  less  than  6  2831920, 
ami  must,  therefore,  be  nearly  equal  half  their  sum  or  a  mean  bo* 
tween  them,  or  62831854,  which,  in  fact,  i^  true  to  the  lasl  figure, 
which  should  be  a  3  instead  of  the  4. 


10  MENSURATION. 

By  the  first  rule,  as  7  :  22 : :  20 :  62f ,  the  answer. 

Ex.  2.  If  the  circumference  of  the  earth  be  25,000 
miles,  what  is  its  diameter? 

By  the  2d  rule,  as  3.1416 : 1 : :  25000 :  7957?,  near- 
ly the  diameter. 

Carol.  To  find  the  circumference  from  the  radius, 
multiply  the  latter  by  6-2832. 

PROBLEM  VIII. 

To  find  the  length  of  any  arc  of  a  circle. 

Multiply  the  degrees  in  the  given  arc  by  the  radius 
of  the  circle,  and  the  product,  again,  by  the  decimal 
•01745,  for  the  length  of  the  arc* 

Ex.  1.  To  find  the  length  of  an  arc  of  30  degrees, 
the  radius  being  9  feet.  Ans.  4-7115. 

"  Ex.  2.  To  find  the  length  of  an  arc  of  12°  10',  or 
12°-},  the  radius  being  10  feet.  Ans.  2-1231. 

PROBLEM  IX. 

To  find  the  area  of  a  circle. 

Rule  I.f  Multiply  half  the  circumference  by  half 
the  diameter.  Or  multiply  the  whole  circumference 
by  the  whole  diameter,  and  take  \  of  the  product. 

Hence  the  circumference  being  6-2831854  when  the  diameter  is  2, 
it  will  be  the  half  of  that,  or  3-1415927,  when  the  diameter  is  1  (th.' 
71),  to  which  the  ratio  in  the  rule,  viz.,  1  to  3-1416,  is  very  near.  Also, 
the  other  ratio  in  the  rule,  7  to  22,  or  1  to  3  j  =  3*1428,  &c.,  is  another 
near  approximation. 

*  It  having  been  found,  in  the  demonstration  of  the  foregoing  prob- 
lem, that  when  the  radius  of  a  circle  is  1,  the  length  of  the  whole 
circumference  is  6-2831854,  which  consists  of  360  degrees ;  therefore, 
as  360° :  6-2831854  : :  1°  :  -01745,  &c.,  the  length  of  the  arc  of  1  de- 
gree. Hence  the  number  -01745,  multiplied  by  any  number  of 
degrees,  will  give  the  length  of  the  arc  of  those  degrees.  And,  be- 
cause the  circumferences  and  arcs  are  as  the  diameters,  or  radii  of  the 
circles ;  therefore,  as  the  radius  1  is  to  any  other  radius  r,  so  is  the 
length  of  the  arc  above  mentioned  to  r  X  -01745  X  degrees  in  the  arc, 
which  is  the  length  of  that  arc,  as  in  the  rule. 

t  This  first  rule  is  proved  in  the  Geometry,  theor.  73. 

And  the  second  rule  is  derived  from  formula  (3),  schol.,  of  the 
same  theorem,  tt  being  3-1416.  Rule  3  is  derived  from  the  same 
formula,  observing  that  7ir"  =.  indr,  and  {it  s=  -7854. 


MENSURATION    OF    PLANES.  11 

Rule  II.  Square  the  radius,  and  multiply  that  square 
by  3- 14 It). 

•  Rule  III.  Square  the  diameter,  and  multiply  that 
square  by  the  decimal  *7854,  for  the  area. 

Ex.  1.  To  find  the  area  of  a  circle  whose  diameter 
is  10,  and  its  circumference  31*416. 

By  Rule  1.  By  Rule  3. 

31-416  -7854 

10  100=  103 

4)31416  78*54 

78*54,  the  area. 
Ex.  2.  To  find  the  area  of  a  circle  whose  diame- 
ter is  7,  and  circumference  22.  Ans.  38|. 

Ex.  3.  How  many  square  yards  are  in  a  circle 
whose  diameter  is  3£  feet?  Ans.  1*069. 

Ex.  4.  Find  the  area  of  a  circle  whose  radius  is  10. 

Ans.  314'] 6. 

PROBLEM  X. 

To  find  the  area  of  a  circular  ring  or  space  included 
between  two  concentric  circles. 

Take  the  difference  between  the  areas  of  the  two 
circles,  as  found  by  the  last  problem.  Or,  since  cir- 
cles are  as  the  squares  of  their  diameters,  subtract 
the  square  of  the  less  diameter  from  the  square  of  the 
greater,  and  multiply  their  difference  by  *7854.  Or, 
lastly,  multiply  the  sum  of  the  diameters  by  the  dif- 
ference of  the  same,  and  that  product  by  *7854  ;* 
which  is  still  the  same  thing,  because  the  product  of 
the  sum  and  difference  of  any  two  quantities  is  equal 
to  the  difference  of  their  squares. 

Ex.  1.  The  diameters  of  two  concentric  circles 
being  10  and  6,  required  the  area  of  the  ring  contained 
between  their  circumferences. 

Here  10  +  6  =  16,  the  sum  ;  and  10  —  6  =  4,  the 
difference. 

Therefore  *7854  x  16  x  4  =7854  x  64  =  50*2656, 
the  area. 

*  In  this  last  method  logarithms  may  be  advantageously  applied. 


12  MENSURATION. 

Ex.  2.  What  is  the  area  of  the  ring,  the  diameters  of 
whose  bounding  circles  are  10  and  20  ?     Ans.  23562. 

PROBLExM   XI. 

To  find  the  area  of  the  sector  of  a  circle. 

Rule  I.  Multiply  the  radius,  or  half  the  diameter, 
by  half  the  arc  of  the  sector,  for  the  area.  Or,  mul- 
tiply the  whole  diameter  by  the  whole  arc  of  the  sec- 
tor, and  take  one  quarter  of  the  product.  The  reason 
for  which  is,  that  the  sector  bears  the  same  propor- 
tion to  the  whole  circle  that  its  arc  does  to  the  whole 
circumference. 

Rule  II.  As  360  is  to  the  degrees  in  the  arc  of  the 
sector,  so  is  the  area  of  the  whole  circle  to  the  area 
of  the  sector. 

This  is  evident,  because  the  sector  is  proportional 
to  the  length  of  the  arc,  or  to  the  degrees  contained 
in  it. 

Ex.  1.  To  find  the  area  of  a  circular  sector  whose 
arc  contains  18  degrees,  the  diameter  being  3  feet. 

1.  By  the  1st  Rule. 

First,  3*1416  X  3  =  9*4248,  the  circumference. 
And  360  :  18  : :  9*4248  :  -47124,  the  length  of  the  arc. 
Then    *47124  x  3  ~  4=  -11781  x  3=35343,    the 
area. 

2.  By  the  2d  Rule. 

First,  -7854  x  32  =  7*0686,  the  area  of  the  whole 
circle. 

Then,  as  360  :  18  : :  7-0686 :  -35343,  the  area  of  the 
sector. 

Ex.  2.  To  find  the  area  of  a  sector  whose  radius 
is  10,  and  arc  20.  Ans.  100. 

Ex.  3.  Required  the  area  of  a  sector  whose  radius 
is  25,  and  its  arc  containing  147°  29'. 

Ans.  804-4017. 

PROBLEM  XII. 

To  find  the  area  of  a  segment  of  a  circle. 
Rule  I.  Find  the  area  of  the  sector  having  the 
same  arc  with  the  segment,  by  the  last  problem. 


MENSURATION    OF   PLANES.  13 

Find,  also,  the  area  of  the  triangle  formed  by  the 
chord  of  the  segment  and  the  two  radii  of  the  sector. 

Then  take  the  sum  of  these  two  for  the  answer, 
when  the  segment  is  greater  than  a  semicircle:  or 
take  their  difference  for  the  answer,  when  it  is  less 
than  a  semicircle ;  as  is  evident  by  inspection. 

Ex.  1.  To  find  the  area  of  the  segment  ACBDA, 
its  chord  AB  being  12,  and  the  radius  AE  or  CE  10. 

*First,  AD  ■£-  AE  =  sin.  angle  D  =  sin.     .      c     B 
36°  524  =  36*87  degrees,  the  degrees  in    /\^~J\ 
the  angle  AEC  or  arc  AC.     Their  double,  /      \:/     \ 
73*74,  are  the  degrees  in  the  whole  arc  I      £!       ; 
ACB.  V     i    S 

Now    -7854x400  =  314-16,  the    area  f* 

of  the  whole  circle. 

Therefore  360°  :  73-74: :  314*16:  64-3504,  area  of 
the  whole  sector  ACBE. 

Again,  v'AE2  —  ADa  =  V 100  —  36  =  %/64  =  8  = 
DE. 

Therefore,  AD  x  DE  =  6  x  8=48,  the  area  of  the 
triangle  AEB. 

Hence  sector  ACBE  —  triangle  AEB  =16-3504, 
area  of  seg.  ACBDA. 

Rule  II.  Divide  the  height  of  the  segment  by  the 
diameter,  and  find  the  quotient  in  the  column  of 
heights  in  the  following  tablet:  Take  out  the  corre- 
sponding area  in  the  next  column  on  the  right  hand, 
and  multiply  it  by  the  square  of  the  circle's  diameter, 
for  the  area  of  the  segment.f 

Note.  When  the  quotient  is  not  found  exactly  in  the 

*  This  requires  a  knowledge  of  plane  trigonometry. 

t  The  truth  of  this  rule  depends  on  the  principle  of  similar  plane 
figures,  which  have  the  ratio  of  their  like  lines  (as  the  height  and  ra- 
dius of  a  segment) equal,  and  are  to  one  another  as  the  square  of  their 
like  linear  dimensions.  The  segments  in  the  table  are  those  of  a  cir- 
cle whose  diameter  is  1 ;  and  the  first  column  contains  the  quotients 
of  corresponding  heights,  or  versed  sines,  divided  by  the  diameter, 
which  are  the  same  for  similar  segments  of  all  diameters.  Thus, 
then,  the  area  of  the  similar  segment,  taken  from  the  table,  and  mul- 
tiplied by  the  square  of  the  diameter,  gives  the  area  of  the  segment 
corresponding  to  this  diameter. 


14 


MENSURATION. 


table,  proportion  may  be  made  between  the  next  less 
and  greater  area,  in  the  same  manner  as  is  done  for 
logarithms  or  any  other  table. 

TABLE  OF   THE   AREAS   OF  CIRCULAR   SEGMENTS. 


1 

+3 

<gg 

i 

i 

•a 
1 

J* 

i 

n 

3 

4-S 

If 

^   0) 

1 

Ti 

-3 

ITT 

1 

1 

•01 

•00133 

•04701 

•11990 

•31 

•20738 

•41 

•30319 

•02 

•00375 

•12 

•05339 

•22 

•12811 

•32 

•21667 

•42 

•31304 

•03 

•00687 

•13 

•06000 

•23 

•13646 

-33 

•22603 

•43 

•32293 

■04 

•01054 

•14 

•06683 

•24 

•14494 

•31 

•23547 

•44 

•33284 

05 

•01468 

•15 

•07387 

•25 

•15354; 

•35 

•24498 

•45 

•34278 

•06 

•01924 

•16 

•08111 

•26 

•16226 

•36 

•25455 

•46 

•35274 

•07 

•02417 

•17 

•08853 

•27 

•17109 

•37 

•26418 

•47 

•36272 

•08 

•02944 

•18 

•09613 

•28 

•18002 

■38 

•27386 

•48 

•37270 

•09 

•03502 

19 

•10390 

•29 

•18905 

•39 

•28359 

•49 

•38270 

•10 

04088 

•20  -11182  1-30 

•19817 

•K) 

•29337 

•50 

•39270 

Ex.  2.  Taking  the  same  example  as  before,  in 
which  are  given  the  chord  AB  12,  and  the  radius  10, 
or  diameter  20. 

And  having  found,  as  above,  DE  =  8  ;  then  CE  — 
DE  =  CD  =  10  —  8  =  2.  Hence,  by  the  rule.  CD  * 
CF  =  2-H20=-1,  the  tabular  height.  This  being 
found  in  the  first  column  of  the  table,  the  correspond- 
ing tabular  area  is  -04088.  Then  :04088  x  202  = 
•04088  x  400=  16-352,  the  area,  nearly  the  same  as 
before. 

Ex.  3.  What  is  the  area  of  the  segment  whose 
height  is  18,  and  diameter  of  the  circle  50  ? 

Ans.  636-375. 

Ex.  4.  Required  the  area  of  the  segment  whose 
chord  is  16,  the  diameter  being  20. 

Ans.  44-7292. 


PROBLEM  XIII. 


To  measure  long  irregular  figures. 

Take  or  measure  the  breadth  in  several  places  at 
equal  distances  ;  then  add  all  these  breadths  together, 


MENSURATION    OF   PLANES.  15 

and  divide  the  sum  by  the  number  of  them  for  the 
mean  breadth,  which  multiply  by  the  length  for  the 
area.* 

Note  1.  Take  half  the  sum  of  the  extreme  breadths 
for  one  of  the  said  breadths. 

Note  2.  If  the  perpendiculars  or  breadths  be  not  at 
equal  distances,  compute  all  the  parts  separately  as 
so  many  trapezoids,  and  add  them  all  together  for  the 
whole  area. 

Or  else  add  all  the  perpendicular  breadths  togeth- 
er, and  divide  their  sum  by  the  number  of  them  for 
the  mean  breadth,  to  multiply  by  the  length ;  which 
will  give  the  whole  area  not  far  from  the  truth. 

Ex.  1.  The  breadths  of  an  irregular  figure,  at  five 
equidistant  places,  being  8*2,  7*4,  9*2,  10*2,  8*6 ;  and 
the  whole  length  39 :  required  the  area. 

First,  (8-2  +  8-6)  ^2  =  8*4,  the  mean  of  the  two 
extremes. 

Then  8*4  +  7.4  +  9*2  +  10*2  =  35*2,  sum  of 
breadths. 

And    35*2  -r    4  =  8*8,  the  mean  breadth. 

Hence  8-8,  x  39  =  343*2,  the  answer. 

Ex.  2.  The  length  of  an  irregular  figure  being 
84,  and  the  breadths  at  six  equidistant  places,  17*4, 
20*6,  14-2,  16*5,  20*1,  24*4  ;  what  is  the  area? 

Ans.  1550*64. 

*  This  rule  is  made  out  as  follows :  Let 
ABCD  be  the  irregular  piece,  having  the 
several  breadths  AD,  EF,  GH,  IK,  BC  at  the 
equal  distances  AE,  EG,  GI,  IB.  Let  the 
several  breadths  in  order  be  denoted  by  the 
corresponding  letters  a,  b,  c,  d,  e,  and  the 
whole  length  AB  by  /;  then  compute  the  areas  of  the  parts  into  which 
tin-  figure  is  divided  by  the  perpendiculars,  as  so  many  trapezoids  by 
Problem  3,  and  add  them  all  together.     Thus  the  sum  of  the  parts  is, 

2±-6  X  AE  +  *±5  X  EG  +  ^B  x  OI+±H  x  IB  , 

=  ( \a  -f  b  +  c  -f  J-f  he)  X  J/  =  (m  -f  b  +  c  +  d)  \l, 
which  is  the  whole  area,  agreeing  with  the  rule  ;  m  being  the  arith- 
metic mean  between  the  extremes  and  4  the  number  of  the  parts. 
And  the  same  for  any  other  number  of  parts. 


MENSURATION  OF  SOLIDS. 


By  the  Mensuration  of  Solids  are  determined  the 
spaces  included  by  contiguous  surfaces;  and  the  sum 
of  the  measures  of  these  including  surfaces  is  the 
whole  Surface  or  Superficies  of  the  body. 

The  measure  of  a  solid  is  called  its  solidity,  capac- 
ity, or  content.     A  better  term  is  volume. 

Solids  are  measured  by  cubes,  whose  sides  are 
inches,  or  feet,  or  yards,  &c.  And  hence  the  volume 
of  a  body  is  said  to  be  so  many  cubic  inches,  feet, 
yards,  &c,  as  will  fill  its  capacity  or  space,  or  anoth- 
er of  equal  magnitude. 

The  least  ordinary  solid  measure,  or  measure  of 
volume,  is  the  cubic  inch,  other  cubes  being  taken 
from  it  according  to  the  proportion  in  the  following 
table: 

Table  of  Cubic  or  Solid  Measures. 

1728  cubic  inches  make  .  .  1  cubic  foot. 

27  cubic  feet  make       .  .  1  cubic  yard. 

166f  cubic  yards  make    .  .  1  cubic  pole. 

64000  cubic  poles  make     .  .  1  cubic  furlong. 

512  cubic  furlongs  make  .  1  cubic  mile. 

PROBLEM  I. 

To  find  the  superficies  of  a  prism. 

Multiply  the  perimeter  of  one  end  of  the  prism  by 
the  altitude  of  one  of  the  parallelograms,  and  the 
product  will  be  the  lateral  surface.  To  which  add, 
also,  the  area  of  the  two  ends  of  the  prism,  when  re- 
quired.* 

*  And  the  rule  is  evidently  the  same  for  the  surface  of  a  cylinder, 
which  may  be  regarded  as  a  prism  of  an  infinite  number  of  lateral 
faces. 


2  MENSURATION. 

Or,  compute  the  areas  of  all  the  sides  and  ends  sep- 
arately, and  add  them  all  together. 

Ex.  1.  To  find  the  surface  of  a  cube,  the  length  of 
each  side  being  20  feet.  Ans.  2400  feet. 

Ex.  2.  To  find  the  whole  surface  of  a  triangular 
prism  whose  length  is  20  feet  and  each  side  of  its  end 
or  base  18  inches.  Ans.  91*948  feet. 

Ex.  3.  To  find  the  convex  surface  of  a  round  prism, 
or  cylinder,  whose  length  is  20  feet  and  diameter  of 
its  base  is  2  feet.  Ans.  125*664. 

Ex.  4.  What  must  be  paid  for  lining  a  rectangular 
cistern  with  lead  at  2d.  a  pound  weight,  the  thickness 
of  the  lead  being  such  as  to  weigh  7  lbs.  for  each 
square  foot  of  surface ;  the  inside  dimensions  of  the 
cistern  being  as  follows,  viz.,  the  length  3  feet  2  inches, 
the  breadth  2  feet  8  inches,  and  depth  2  feet  6  inches  ? 

Ans.  £2  3s.  10  ±d. 

To  find  the  superficies  of  an  irregular  polyhedron. 

Find  the  superficies  of  each  of  its  bounding  polyg- 
onal faces,  and  add  the  results. 

To  find  the  superficies  of  a  regular  polyhedron. 

Find  the  area  of  one  of  its  faces  by  Prob.  VI.,  and 
multiply  this  by  the  number  of  faces. 

PROBLEM    II. 

To  find  the  surface  of  a  regular  pyramid  or  cone. 

Multiply  the  perimeter  of  the  base  by  the  slant 
height,  or  length  of  the  side,  and  half  the  product 
will  evidently  be  the  convex  surface  or  the  sum  of 
the  areas  of  all  the  triangles  which  form  it.  To 
which  add  the  area  of  the  end  or  base,  if  requisite. 
Note.  The  slant  height  of  a  regular  pyramid  is  the 
perpendicular  from  the  vertex  to  the  middle  of  one 
of  the  sides  of  the  base. 

Ex.  1.  What  is  the  upright  surface  of  a  triangular 
pyramid,  the  slant  height  being  20  feet,  and  each  side 
of  the  base  3  feet  ?  Ans.  90  feet. 


MENSURATION    OF   SOLIDS.  J 

Ex.  2.  Required  the  convex  surface  of  a  cone,  or 
circular  pyramid,  the  slant  height  being  50  feet,  and 
the  diameter  of  its  base  8i  feet.  Ans.  667*59. 

PROBLEM  III. 

To  find  the  surface  of  the  frustum  of  a  regular  'pyr- 
amid or  cone,  being  the  lower  part,  when  the  top  is  cut 
off  by  a  plane  parallel  to  the  base. 

Rule  I.  Add  together  the  perimeters  of  the  two 
ends,  and  multiply  their  sum  by  the  slant  height,  tak- 
ing half  the  product  for  the  answer.  Because  the 
lateral  faces  of  the  frustum  of  a  pyramid  are  trape- 
zoids, having  their  opposite  sides  parallel,  and  the 
frustum  of  a  cone  is  the  frustum  of  a  pyramid  of  an 
infinite  number  of  lateral  faces. 

Rule  II.  Multiply  the  perimeter  of  the  section  mid- 
way between  the  two  bases  by  the  slant  height. 
This  depends  upon  the  fact  that  the  perimeter  of  the 
middle  section  is  half  the  sum  of  the  perimeters  of 
the  bases,  as  may  be  easily  shown. 

Ex.  1.  How  many  square  feet  are  in  the  surface 
of  the  frustum  of  a  square  pyramid  whose  slant 
height  is  10  feet;  also,  each  side  of  the  base  or  great- 
er end  being  3  feet  4  inches,  and  each  side  of  the  less 
end  2  feet  2  inches?  Ans.  110  feet. 

Ex.  2.  To  find  the  convex  surface  of  the  frustum 
of  a  cone,  the  slant  height  of  the  frustum  being  12| 
feet,  and  the  circumferences  of  the  two  ends  6  and 
8-4.  Ans.  90  feet. 

PROBLEM    IV. 

To  find  the  volume  of  any  prism  or  cylinder. 
Find  the  area  of  the  base,  or  end,  whatever  the  fig- 
ure of  it  may  be ;  and  multiply  it  by  the  altitude*  of 
the  prism,  or  cylinder,  for  the  volume. 

*  The  altitude  of  a  prism  is  the  perpendicular  distance  between  its 
parallel  bases.  The  cylinder,  as  well  as  the  prism,  may  be  oblique. 
Prop.  3  of  Solid  Geom.,  upon  which,  with  the  note  to  Prop.  6  of  the 
6ame.  the  demonstration  of  thi9  depends,  may  evidently  be  extended 
to  an  oblique  cylinder. 


4  MEXSI   RATION. 

Ex.  1.  Find  the  solid  content  of  a  cube  whose  side 
is  24  inches.  Ans.  13824. 

Ex.  2.  How  many  cubic  feet  are  in  a  block  of  mar- 
ble, its  length  being  3  feet  2  inches,  breadth  2  feet  8 
inches,  and  thickness  2  feet  6  inches?      Ans.  2\~. 

Ex.  3.  How  many  gallons  of  water  will  the  cis- 
tern contain  whose  dimensions  are  the  same  as  in  the 
last  example,  when  277*274  cubic  inches  are  contained 
in  one  gallon?  Ans.  131*566. 

Ex.  4.  Required  the  solidity  of  a  triangular  prism 
whose  length  is  10  feet,  and  the  three  sides  of  its  tri- 
angular end  or  base  are  3,  4,  5  feet.  Ans.  60. 

Ex.  5.  Required  the  content  of  a  round  pillar,  or 
cylinder,  whose  length  is  20  feet,  and  circumference 

5  feet  6  inches.  Ans.  48-1459. 

PROBLEM    V. 

To  find  the  volume  of  any  pyramid  or  cone. 

Find  the  area  of  the  base,  and  multiply  that  area 
by  the  perpendicular  height ;  then  take  one  third  of 
the  product  for  the  volume.  (See  Prop.  XL,  Solid 
Geom.,  and  corollaries.) 

Ex.  1.  Required  the  solidity  of  the  square  pyramid, 
each  side  of  its  base  being  30,  and  its  perpendicular 
height  25.  Ans.  7500. 

Ex.  2.  To  find  the  content  of  a  triangular  pyramid 
whose  perpendicular  height  is  30,  and  each  side  of 
the  base  3.  Ans.  38-97117. 

Ex.  3.  To  find  the  content  of  a  triangular  pyramid, 
its  height  being  14  feet  6  inches,  and  the  three  sides 
of  its  base  5,  6,  7.  Ans.  71-0352. 

Ex.  4.  What  is  the  content  of  a  pentagonal  pyra- 
mid, its  height  being  12  feet,  and  each  side  of  its  base 
2  feet  ?  Ans.  27*5276. 

Ex.  5.  What  is  the  content  of  the  hexagonal  pyr- 
amid whose  height  is  6-4,  and  each  side  of  its  base 

6  inches?  Ans.   1*38564  feet. 

Ex.  6.  Required  the  content  of  a  cone,  its  height 
being  10|  feet,  and  the  circumference  of  its  base  9 
feet.  Ans.   22*56093. 


MENSURATION    OF    SOLIDS. 


PROBLEM   VI. 

To  find  the  volume  of  the  frustum  of  a  cone  or  pyramid. 

Rule  I.  Add  into  one  sum  the  areas  of  the  two 
ends,  and  the  mean  proportional  between  them,  or 
the  square  root  of  the  product,  and  one  third  of 
that  sum  will  be  -a  mean  area ;  which,  being  multi- 
plied by  the  perpendicular  height  or  length  of  the 
frustum,  will  give  its  content. 

Rule  II.  For  the  cone.  Add  together  the  squares 
of  the  radii  of  the  two  bases  and  their  product,  and 
multiply  the  sum  by  3*1416,  and  the  product  by  one 
third  of  the  altitude.  (See  Prop.  XII.,  Solid  Geom., 
and  corol.) 

Ex.  1.  To  find  the  number  of  solid  feet  in  a  piece 
of  timber  whose  bases  are  squares,  each  side  of  the 
greater  end  being  15  inches,  and  each  side  of  the 
less  end  6  inches ;  also,  the  length  or  perpendicular 
altitude  24  feet  ?  Ans.  19 J. 

Ex.  2.  Required  the  content  of  a  pentagonal  frus- 
tum whose  height  is  5  feet,  each  side  of  the  base  18 
inches,  and  each  side  of  the  top  or  less  end  6  inches. 

Ans.  931925  feet. 

Ex.  3.  To  find  the  content  of  a  conic  frustum,  the 
altitude  being  18,  the  greatest  diameter  8,  and  the 
least  diameter  4.  Ans.  527-7888. 

Ex.  4.  What  is  the  volume  of  the  frustum  of  a 
cone,  the  altitude  being  25 ;  also,  the  circumference 
at  the  greater  end  being  20,  and  at  the  less  end  10  ? 

Ans.  464216. 

Ex.  5.  If  a  cask,  which  is  two  equal  conic  frus- 
tums joined  together  at  the  bases,  have  its  bung  di- 
ameter 28  inches,  the  head  diameter  20  inches,  and 
length  40  inches,  how  many  gallons  of  wine  will  it 
hold?  Ans.  79*0613. 

PROBLEM  VII. 

To  find  the  surface  of  a  sphere,  or  any  segment. 
Rule  I.  Multiply  the  circumference  of  the  sphere 


c, 


MENSURATION. 


by  its  diameter,  and  the  product  will  be  the  whole 
surface  of  it.* 

Rule  II.  Multiply  the  square  of  the  diameter  by 
3-1416,  and  the  product  will  be  the  surface. 

Note.  For  the  surface  of  a  segment,  multiply  the 
circumference  of  a  great  circle  of  the  sphere  by  the 
altitude  of  the  segment. 

Ex.  1.  Required  the  convex  superficies  of  a  sphere 
whose  diameter  is  7,  and  circumference  22. 

Ans.  154. 


*  For  if  a  regular  semi-polygon  be  revolved 
about  a  diameter  of  the  figure,  each  of  the  trap- 
ezoids, as  BGHC,  will  describe  the  frustum  of 
a  cone,  the  convex  surface  of  which  will*  be 
measured  by  the  circumference  of  MN,  describ- 
ed by  the  middle  point  of  its  inclined  side,  multi- 
plied by  the  slant  height  BC  (Prob.  3,  Rule  2). 
But  by  the  similarity  of  the  triangles  IMN  and 
BCO,  whose  sides  are  respectively  perpendicular, 

BC  :  BO  : :  IM  :  MN  :  circum.  IM :  circum.  MN 
(Geom.,  th.  71). 

.-.  BC  X  circum.  MN  =  BO  X  circum.  IM. 

In  the  same  manner,  the  convex  surface  of  the 
frustum  described  by  the  revolution  of  the  trap- 
ezoid HCDK  may  be  shown  to  be  measured  by 
HK  X  circum.  IM.  Of  that  described  by  the 
revolution  of  DKLE  by  KL  X  circum.  IM.  And,  by  addition,  the 
surface  described  by  the  portion  of  the  perimeter  BCDE  is  measured 
by  GL  X  circum.  IM.  The  same  result  will  be  obtained  when  the 
number  of  sides  of  the  semi-polygon  is  infinite  and  it  becomes  a  semi- 
circle, generating  a  sphere  by  its  revolution  ;  and  the  portion  BCDE 
generating  a  zone,  of  which  GL  is  the  altitude.  The  circum.  IM  in 
this  case  becomes  the  circumference  of  a  great  circle  of  the  sphere. 
When  the  whole  semi-polygon  or  semicircle  revolves,  the  altitude  be- 
comes the  diameter  AF,  and  the  surface  is  measured  by  the  circum- 
ference of  a  great  circle  multiplied  by  its  diameter.  This  is  equal  to 
four  times  the  area  of  a  great  circle  (see  th.  73,  Geom.). 

Corol.  The  convex  surface  of  a  cylinder  circumscribing  a  sphere  is 
measured  by  the  rectangle  of  the  circumference  of  the  base  by  the 
altitude,  which,  being  equal  to  the  diameter  of  the  sphere,  and  the 
base  of  the  cylinder  equal  a  great  circle,  it  follows  that  the  measure 
of  the  surface  of  the  sphere  is  equal  to  that  of  the  convex  surface  of 
the  cylinder.  If  now  we  add  the  two  bases  of  the  cylinder,  since  the 
surface  of  the  sphere  is  equal  to  four  great  circles,  we  shall  have  the 
surface  of  the  cylinder  equal  six  great  circles,  so  that  the  surfaces  of 
the  sphere  and  circumscribed  cylinder  are  as  4  to  6,  or  as  2  to  3. 
Rule  2  follows  obviously  from  Rule  1. 


MENSURATION    OF   SOLIDS.  7 

Ex.  2.  Required  the  superficies  of  a  globe  whose 
diameter  is  24  inches.  Ans.  1809-5G1G. 

Ex.  3.  Required  the  area  of  the  whole  surface  of 
the  earth,  its  diameter  being  7957f  miles,  and  its  cir- 
cumference 25000  miles. 

Ans.  198943750  sq.  miles. 

Ex.  4.  The  axis  of  a  sphere  being  42  inches,  what 
is  the  convex  superficies  of  the  segment  whose  height 
is  9  inches?  Ans.  1187*5248  inches. 

Ex.  5.  Required  the  convex  surface  of  a  spherical 
zone  whose  breadth  or  height  is  2  feet,  and  cut  from 
a  sphere  of  12£  feet  diameter.         Ans.  78*54  feet. 

PROBLEM  VIII. 

To  find  the  surface  of  a  lune. 

Multiply  the  arc  which  measures  the  angle  of  the 
lune  by  the  diameter  of  the  sphere.  For  the  lune  is 
to  the  whole  surface  of  the  sphere  as  its  arc  is  to  a 
circumference. 

Cor.  1.  The  measure  of  a  spherical  wedge,  or  un- 
gula,  is  for  a  similar  reason  the  product  of  the  lune 
which  serves  for  its  base,  multiplied  by  one  third  the 
radius  of  the  sphere  (see  next  Prob.). 

Cor.  2.  The  measure  of  a  spherical  triangle  is  the 
arc  of  a  great  circle  subtending  half  the  excess  of  the 
sum  of  its  angles  over  two  right  angles,  multiplied  by 
the  diameter  of  the  sphere.  This  depends  on  the 
above  and  Prop.  XVII,,  cor.  l,Spher.  Geom. 

The  measure  of  the  surface  of  a  spherical  polygon 
is  the  arc  of  a  great  circle  subtending  half  the  excess 
of  the  sum  of  its  angles  over  as  many  times  two 
right  angles  as  the  figure  has  sides,  wanting  two,  mul- 
tiplied by  the  diameter  of  the  sphere.* 

Or  in  symbols,  s  denoting  the  sum  of  the  angles  of 
the  polygon  in  fractions  of  a  right  angle,  n  the  num- 
ber of  its  sides, 

[s  —  2(ti  —  2)]  H-  8, 

*  This  may  be  easily  proved  by  dividing  the  polygon  ruto  tri- 


8  MENSURATION. 

will  express  the  fraction  which  the  polygon  is  of  the 
whole  surface  of  the  sphere. 

Ex.  1.  Required  the  surface  of  the  lune  whose  arc 
is  8  and  diameter  10.  Ans.  8  X  10  =  80. 

Ex.  2.  Required  the  measure  of  the  lune  whose 
angle  is  30°  20',  and  diameter  12. 

Ans.   12  X  3-1416  x  12  X  ^'  =38-139.* 

Ex.  3.  Required  the  area  of  a  spherical  triangle, 
of  which  the  three  angles  are  30°,  100°,  and  80°,  the 
diameter  of  the  sphere  being  40. 

Ans.  ~  X  3-1416  x  40X40. 

Ex.  4.  Required  the  area  of  a  spherical  pentagon, 
the  angles  of  which  are  60°,  110°,  150°,  160°,  and 
100°,  the  diameter  of  the  sphere  being  50. 

Ans.  ^[60  +  110+150  +  160  +  100  — (5— 2)180°] 
-f-360}X  3-1416X50X50. 

Ex.  5.  What  fraction  of  the  whole  surface  of  a 
sphere  is  a  spherical  heptagon,  the  angles  of  which 
are  in  fractions  of  a  right  angle  1J,  1},  If,  li,  If, 

Ans   "l-M-is-i-fl-11 
Ans.      g      — 18  .  o  —  -. 


PROBLEM  IX. 

To  find  the  volume  of  a  sphere  or  globe. 

.  Rule  I.  Multiply  the  surface  by  the  diameter,  and 
take  one  sixth  of  the  product  for  the  content. 

Rule  II.  Multiply  the  cube  of  the  diameter  by  the 
decimal  -5236  for  the  content. 

Ex.  1.  To  find  the  content  of  a  sphere  whose  axis 
is  12.  Ans.  904*7808. 

Ex.  2.  To  find  the  solid  content  of  the  globe  of  the 
earth,  supposing  its  circumference  to  be  25,000  miles. 
Ans.  263,857,437,760  miles. 

*  This  answer  is  of  the  same  denomination  as  the  diameter,  ex- 
cept that  it  is  square  units  instead  of  linear.  Logarithms  may  here 
be  conveniently  applied,  using  the  arithmetical  complement  of  the 
logarithm  of  the  divisor  360°  reduced  to  minutes. 


MENSURATION    OF    SOLIDS.  V 

PROBLEM   X. 

To  find  the  volume  of  a  spherical  sector. 

Multiply  the  area  of  the  zone  which  serves  for  its 
base  by  one  third  of  the  radius  of  the  sphere.* 

Ex.  1.  Required  the  volume  of  a  spherical  sector, 
the  altitude  of  the  zone  which  serves  for  a  base  being 
12,  and  the  diameter  of  the  sphere  being  30. 

Ans.   12  X  30  X  3-1410  X  5. 

Ex.  2.  Required  the  volume  of  a  spherical  sector, 
a  great  section  of  the  zone  base  being  an  arc  of  40°, 
and  the  diameter  of  the  sphere  being  100. 

PROBLEM    XI. 

To  find  the  volume  of  a  spherical  segment. 

Rule  I.    From   three   times  the   diameter  of  the 
sphere  take  double  the  height  of  the  segment ;  then 
multiply  the  remainder  by  the  square  of  the  height, 
and  the  product  by  the  decimal  -523G  for  the  coi. 
(See  Schol.  to  Prop.  XIV.,  Sol.  Geom.) 

Rule  II.  To  three  times  the  square  of  the  radii; 
the  segment's  base  add  the  square  of  its  height ;  t! 
multiply  the  sum  by  the  height,  and  the  product  by 
•5236,  for  the  content. 

Rule  III.  When  the  segment  has  two  bases,  multi- 
ply the  half  sum  of  the  parallel  bases  by  the  altitude, 
and  add  the  volume  of  the  sphere  of  which  this  alti- 
tude is  the  diameter. 

*  The  spherical  sector  may  be  Apposed  to  he  made  up  of  an  in- 
finite number  of  indefinitely  small  cones, each  taring  an  evai 
portion  of  the  surface  of  the  zone  base  for  a  base,  mid  the  radius  of 
the  sphere  for  an  altitude.  The  sum  of"  these  will  he  measured  by 
the  sum  of  their  bases,  or  the  zone  multiplied  by  one  third  their  Com 
mmi  altitude,  or  the  radius  of  the  sphere. 

When  the  zone  becomes  the  whole  surface  of  the  sphere  the  sector 
becomes  the  whole  solid  sphere.     Note  that  one  third  the  radius  is 
ii  the  diameter. 

Un\<-  ■„'.  observe  that  ir<P  z=  surface  of  sphere  (Prob.  7),  and 
5~:=-">-3G.  The  rule  is  similar  for  a  spherical  pyramid  having  a 
spherical  polygon  fnr  a  base  and  the  center  of  the  sphere  for  a  vertex. 

K 


10  MENSURATION. 

Ex.  1.  To  find  the  content  of  a  spherical  segment 
of  two  feet  in  height  cut  from  a  sphere  of  8  feet  in 
diameter.  Ans.  41*888. 

Ex.  2.  What  is  the  solidity  of  the  segment  of  a 
sphere,  its  height  being  9,  and  the  diameter  of  its  base 
20?  Ans.  1795-4244. 

EXERCISES  IN  MENSURATION.* 

1.  Transform  a  given  parallelogram  into  another  of  double  the  alti- 
tude which  shall  have  a  given  angle. 

2.  To  transform  a  triangle  into  another  of  the  same  base  and  given 
vertical  angle. 

3.  To  construct  a  triangle  of  given  base,  vertical  angle,  and  area. 

4.  The  same,  except  the  altitude  instead  of  the  base  given. 

5.  To  construct  a  triangle  similar  to  a  given  triangle,  and  equal  to 
a  given  square. 

6.  A  triangle  with  given  angles  at  the  base,  and  equal  to  a  given 
rectangle. 

7.  The  same,  when  the  base,  vertical  angle,  and  rectangle  of  the 
other  two  sides  are  given. 

8.  The  same,  when  the  base,  the  altitude,  and  the  product  of  the 
two  sides. 

9.  The  same,  when  the  altitude,  the  area,  and  the  ratio  of  one  of 
the  sides  to  the  base. 

10.  The  same,  when  the  ratio  of  the  base  and  altitude,  the  vertical 
angle  and  the  area. 

11.  Make  a  regular  hexagon  equivalent  to  a  given  polygon. 

12.  To  construct  a  figure  similar  to  a  given  figure,  and  its  area  hav- 
ing to  that  of  the  given  figure  a  given  ratio. 

13.  A  quadrilatei'al  capable  of  being  inscribed,  in  which  two  ad- 
jacent angles,  the  angle  which  its  diagonals  make  with  each  other 
and  its  ai-ea,  are  given. 

14.  A  quadrilateral  that  may  be  inscribed,  in  which  three  angles 
and  the  area  are  given. 

15.  A  circle  equal  to  the  sum  of  several  circles. 

16.  A  square  in  a  given  semicircle. 

17.  A  circle  equal  to  the  ring  between  two  circles. 

18.  A  quadrant  equal  to  a  given  semicircle. 


*  Many  of  these  will  conveniently  admit  the  application  of  loga- 
rithms. 


KXERCISES    IN    MENSURATION.  11 

15).  A  sextant  equal  to  a  given  quadrant. 

2U.  To  determine  the  side  of  an  equilateral  triangle,  the  area  of 
which  is  73-15. 

2 1 .   Also,  of  a  regular  hexagon,  the  area  of  which  is  1G8. 

The  side  of  a  regular  pentagon  is  21-7.     What  is  that  of  another 
half  as  large  ? 

To  find  the  radius  of  a  semicircle  equal  to  a  triangle  whoso 
-  1  !.  and  altitude  9. 
24.   What  18  the  diameter  of  a  circle  equal  to  a  trapezoid,  of  which 
the  base  is  17*4,  the  opposite  side  127,  and  tho  altitude  1008? 

To  find  the  content  of  a  regular  octagon  when  the  radius  of  its 
[bed  circle  is  equal  to  12. 
2G.  Of  a  regular  decagon  when  the  radius  of  the  inscribed  circle  is 
equal  to  17-2. 

27.  How  large  is  the  anglo  at  the  center  of  a  circular  sector,  the 
area  of  which  is  equal  to  that  of  an  equilateral  triangle  whose  side  is 
14,  the  radius  of  the  sector  being  8  ? 

28.  To  determine  the  side  of  a  square  which  shall  be  equal  to  a 
whose  arc  is  18°,  and  radius  7*5. 

29.  To  determine  the  diameter  of  a  circle  which  shall  be  equal  to 
the  segment  of  a  circle  whose  radius  is  120,  and  arc  135°. 

30.  To  determine  the  radii  of  the  inscribed  and  circumscribed  cir- 
cles of  a  triangle  whose  sides  are  10,  12,  and  11. 

31.  To  find  the  convex  surface  of  a  regular  hexagonal  prism,  the 
longest  diameter  of  which  is  2r,  and  height  h. 

Of  a  regular  hexagonal  pyramid  with  the  same  data. 
33.  Of  a  zone  of  one  base,  the  radius  of  which  is  r,  and  altitude  h. 
3  1.  Of  a  spherical  sector,  the  chord  of  which  =c,  and  rad.  sphere 
=  r. 

35.  To  find  the  volume  of  a  solid  generated  by  the  revolution  of 
tor  of  a  circle  about  a  line  through  the  center,  and  exterior  to  the 

36.  Find  the  volume  of  the  solid  generated  by  the  revolution  of  any 
triangle  about  one  of  its  sides. 

37.  Find  the  volume  of  a  solid  generated  by  the  revolution  of  the 
segment  of  a  circle  about  a  line  passing  through  the  center  of  the  cir- 

l  exterior  to  the  segment. 

38.  Prove  that  the  surfaces  of  two  spheres  are  as  the  squares,  and 
the  volumes  as  the  cubes  of  their  radii. 

39.  Find  the  volume  left  of  a  cylinder  after  a  spherical  segment 
having  one  bate  equal  to  that  of  the  cylinder,  and  the  same  altitude 
with  the  cylinder,  has  been  abstracted. 

40.  The  altitude  and  surface  of  a  regular  hexagonal  prism,  of 


12  MENSURATION. 

which  the  greatest  diameter  is  18,  and  the  volume  of  which  is 

to  that  of  a  regular  triangular  pyramid,  of  which  the  is  equal 

to  8,  and  altitude  20. 

41.  In  a  quadrangular  and  hexagonal  prism  each  side  of  the  bases  is 
7,  the  height  13.     What  is  the  ratio  of  their  volumes  and  surfaces? 

42.  To  find  the  altitude  of  a  regular  quadrangular  pyramid,  the 
side  of  whose  base  is  28-7,  and  volume  equal  to  that  of  a  rectangular 
parallelopipedon  whose  edges  are  13,  17,  and  23. 

43.  The  ratio  of  two  homologous  edges  of  two  similar  polvhodrons 
is  5  :  7.     To  find  the  ratio  of  their  surfaces  and  volumes. 

44.  The  ratio  of  the  volumes  of  two  similar  polyhedrons  is  14  :  29. 
To  find  that  of  their  homologous  edges. 

45.  What  is  the  ratio  of  the  surfaces. of  two  regular  pyramids,  the 
one  triangular,  the  other  quadrangular,  if  the  base  in  both  is  24,  and 
the  altitude  7  ? 

46.  A  regular  tetrahedron,  the  edge  of  which  is  15,  has  the  third  of 
its  altitude  cut  off  by  a  plane  parallel  to  the  base ;  required  the  vol- 
ume of  the  frustum  left.     Also,  the  surface. 

47.  About  a  sphere  of  16  inches  radius  a  polyhedron  is  circum- 
scribed, containing  20,800  cubic  inches.  What  is  the  area  of  the 
surface  of  the  latter  1 

48.  A  cylinder  and  cone  have  their  radii  14  and  8,  their  altitudes  6 
and  9.     What  is  the  ratio  of  their  volumes  and  surfaces  ? 

49.  Find  the  radius  of  a  sphere  equal  to  a  cube,  the  diagonal  of 
which  is  17-22. 

50.  Also,  of  a  sphere-  equal  to  a  regular  octahedron,  the  diagonal  of 
which  is  31*5. 

51.  Find  the  radius  of  an  inscribed  sphere  in  a  regular  tetrahedron, 
the  edge  of  which  is  a. 

52.  The  same  for  an  octahedron. 

53.  Find  the  ratio  of  the  surfaces  of  a  regular  tetrahedron  and  in- 
scribed sphere. 

54.  The  same  for  an  octahedron  and  sphere. 

55.  What  is  the  ratio  of  a  hemisphere  to  a  cone  of  the  same  base 
and  altitude  ? 

56.  Find  the  ratio  of  the  solids  generated  by  a  triangle  and  rectan- 
gle revolving  about  a  common  base,  and  the  altitude  of  the  former 
being  double  that  of  the  latter. 

57.  To  find  the  volume  of  a  spherical  segment  when  the  radius  is 
5-86,  and  the  arc  of  a  great  section  162°  14'. 

58.  Find  the  base  of  a  square  pyramid  which  shall  contain  a 
cubic  yard,  and  the  altitude  of  which  shall  be  1  foot. 


EXERCISES    IN   MENSURATION.  13 

The  ride*  of  the  base  of  a  tetrahedron  are  12,  15,  17,  its  alti- 
tude 9.     Required  its  volume.  Ans.  263*248. 

GO.  A  regular  tetrahedron  contains  19*683  cubic  yards.  Required 
its  edges  and  surface.  Ans.  Edge  5-50705,  surfac. 

61.  Required  the  volume  of  a  frustum  of  a  regular  triangular  pyra- 
mid, the  larger  base  of  which  has  0-9  for  its  side,  and  the  smaller 
base  0-4,  and  of  which  the  lateral  edge  is  0-5.  Ans.  0078371. 

I  riven  the  volume  of  a  sphere  equal  to  1843-080278  to  find  its 
radius.  Ans.  7*61. 

63.  Given  the  edge  of  a  cube  0-3G.  Required  the  volume  of  the 
circumscribed  Bphere.  Ans.  0-120937. 

64.  Find  the  area  of  a  spherical  triangle,  the  angles  of  which  are 
tively    85  grades,  17',  T03sr,  35',  67er,  49',  the  radius  of  the 

sphere  being  1*54.  Ans.  2-0805. 

65.  There  is  a  crucible  in  the  form  of  a  conic  frustum,  the  bottom 
of  which  is  003  in  diameter,  the  top  0-00,  and  the  altitude  0-08 ;  this 
crucible  contains  a  quantity  of  melted  metal,  the  surface  of  which  is 
005  in  diameter:  it  is  required  to  make  a  sphere  of  it.  What  is  the 
diameter  of  the  proper  mold  ?  Ans.  0-5074 1 1. 

60.  Given  the  side  or  apophthegm  of  a  cone  25*15,  and  its  height 
17-3,  to  find  its  convex  surface  and  volume. 

Ans.  A  =  1442-32,  V  =  0037-01. 

67.  Find  the  quantity  of  glass  in  a  lens,  of  which  the  diameters  of 
the  surfaces  are  0*03,  and  the  thickness  of  the  lens  0-004.* 

Ans.  0-000001422094. 

68.  Supposing  the  earth  to  be  perfectly  spherical,  and  a  quarter  of 
the  meridian  to  be  expressed  by  10,000,000;  find  me  expression  for  its 
radius,  the  area  of  its  surface,  its  volume  and  weight,  supposing  the 
mean  density  of  the  earth  to  be  5*6604.f 

*  This  solid  is  a  double  segment  of  a  sphere. 

t  This  number  is  the  result  of  the  experiments  of  Sir  Francis 
Bailey,  given  in  the  xivth  vol.  of  the  Memoirs  of  the  Royal  Ast.  Soc. 
of  Lond.,  1844. 


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